1

1. 𝛿(𝑎𝑡) equals |𝑎| 𝛿(𝑡)

Ans b
2. Given the signal 𝑥(𝑡) = 1 − |𝑡| for |𝑡| ≤ 1 and 0 otherwise, which is a triangular pulse,
input to an LTI system with impulse response ℎ(𝑡) = 𝑢(𝑡 + 2) − 𝑢(𝑡 − 2), which is a
rectangular pulse. To evaluate the convolution, we flip 𝑥(𝑡) and shift while evaluating
the area under the product. It can be seen the minimum is 0 and occurs when there is
no overlap between the triangular pulse and the rectangular pulse. The peak is 1, which
occurs when the triangular pulse completely overlaps with the rectangular pulse.
Further, this happens for −1 ≤ 𝑡 ≤ 1
Ans b
sin(𝑎𝑡) sin(𝐹𝑡) sin(𝑎𝑡)
3. Note ↔ 𝑝2𝑎 (𝜔). Given 𝑥(𝑡) = =𝜋 with 𝑎 = 𝐹. The Fourier
𝜋𝑡 𝑡 𝜋𝑡
transform is 𝑋(𝜔) = 𝜋𝑝2𝐹 (𝜔). Hence, using Parseval’s relation, energy is
∞
1 1
∫ |𝑋(𝜔)|2 𝑑𝜔 = × 𝜋 2 × 2𝐹 = 𝜋𝐹
2𝜋 2𝜋
−∞
𝐹
sin(𝐹𝑡) sin (𝜋 𝜋 𝑡) 𝐹 𝐹 𝐹
=𝐹 = 𝜋 sinc ( 𝑡) ↔ 𝜋 for |𝑓| ≤ , 0 otherwise
𝑡 𝐹 𝜋 𝜋 2𝜋
𝜋𝜋𝑡
𝐹
Hence, energy is 𝜋 2 × 2 × 2𝜋 = 𝜋𝐹

Ans d
4. This can be simplified as follows

∞ ∞
1 𝑢
∫ sin(𝑢) 𝛿(6𝑢 − 15𝜋)𝑑𝑢 = ∫ sin ( ) 𝛿(𝑢 − 15𝜋)𝑑𝑢
−∞ −∞ 6 6
∞
1 𝑢 + 15𝜋 1 15𝜋 1 5𝜋 1
=∫ sin ( ) 𝛿(𝑢)𝑑𝑢 = sin ( ) = sin ( ) =
−∞ 6 6 6 6 6 2 6

Ans a
∞
5. The value of ∫−∞ 𝜙(𝑡)𝛿 ′ (𝑡) 𝑑𝑡 equals −𝜙 ′ (0). Let 𝜙1 (𝑡) = 𝜙(𝑡 + 𝛼)
∞ ∞
∫ 𝜙(𝑡 + 𝛼)𝛿 ′ (𝑡)
𝑑𝑡 = ∫ 𝜙1 (𝑡)𝛿 ′ (𝑡) 𝑑𝑡 = −𝜙1′ (0) = −𝜙 ′ (𝛼)
−∞ −∞

Ans c
6. An example of a random signal is noise
Ans a
7. The quantity
∞ ∞
𝑡 𝑡 2𝜏
∫ sin ( ) 𝛿 ′ ( ) 𝑑𝑡 = 2 ∫ sin ( ) 𝛿 ′ (𝜏) 𝑑𝜏
−∞ 3 2 −∞ 3
∞
∫−∞ ϕ(𝑡) 𝛿 ′ (𝑡)𝑑𝑡 = −ϕ′ (𝑡)|𝑡=0. Hence,
∞
2𝜏 2𝜏 2 2𝜏 4
2 ∫ sin ( ) 𝛿 ′ (𝜏) 𝑑𝜏 = −2 sin′ ( )| = −2 × cos ( )| =−
−∞ 3 3 𝜏=0 3 3 𝜏=0 3
 Ans a
8𝜋𝑛
8. 𝑥(𝑛) = sin ( 18 ) is a discrete time, power and periodic signal with period 𝑁 = 9. It is
not an energy signal since its energy is infinite
Ans d
sin(𝑎𝑡) sin(𝐹𝑡) sin(𝑎𝑡)
9. Note ↔ 𝑝2𝑎 (𝜔). Given 𝑥(𝑡) = =𝜋 with 𝑎 = 𝐹. The Fourier
𝜋𝑡 𝑡 𝜋𝑡
transform is 𝑋(𝜔) = 𝜋𝑝2𝐹 (𝜔). Hence, using Parseval’s relation, energy is
∞
1 1
∫ |𝑋(𝜔)|2 𝑑𝜔 = × 𝜋 2 × 2𝐹 = 𝜋𝐹
2𝜋 2𝜋
−∞
1 sin(8𝑡) 1
Therefore, energy of 2 is 4 × 𝜋8 = 2𝜋
𝑡

Ans c
∞
10. Area under impulse is ∫−∞ 𝛿(𝑡)𝑑𝑡 = 1
Ans d
1. An eigenfunction of the LTI system is 𝑒 𝑗2𝜋𝑓0𝑡
Ans d
3𝜋 5𝜋
2. Given 𝑒 𝑗 7 𝑡 + 𝑒 𝑗 9 𝑡 .
3𝜋 3𝜋
(𝑡+𝑇1 ) 7 14𝐾1
𝑒𝑗 7 𝑡 = 𝑒𝑗 7 ⇒ 𝑇1 = 𝐾1 2𝜋 × =
3𝜋 3
18𝐾2
𝑇2 =
5
14𝐾1 18𝐾2 𝐾1 27
𝑇1 = 𝑇2 = = ⇒ =
3 5 𝐾2 35
Setting 𝐾1 = 27, 𝐾2 = 35 yields 𝑇1 = 𝑇2 = 126
Therefore, fundamental period is 126
Ans b
3. Given signal
1 −1 ≤ 𝑡 < 0
𝑥(𝑡) = {−1 0 ≤ 𝑡 < 1
0 otherwise
given as input to LTI system with impulse response ℎ(𝑡) = 1 for 0 ≤ 𝑡 ≤ 1 and 0
otherwise. Resulting output 𝑦(𝑡) is

𝑡 + 1 −1 ≤ 𝑡 < 0
1 − 2𝑡 0 ≤ 𝑡 < 1
𝑦(𝑡) = 𝑡−2 1≤𝑡 <2
0 𝑡 < −1
{ 0 𝑡>2
𝑑𝑦(𝑡)
Hence, it can be seen that the only true statement is that 𝑑𝑡 = −2 for 0 ≤ 𝑡 < 1
Ans b
4. The modulator is linear but not time-invariant
Ans c
5. A linear time-invariant (LTI) system has to satisfy additivity, homogeneity and time-
invariance properties
Ans b
6. Given input 𝑥(𝑛) = 𝛼 𝑛 𝑢(𝑛) given to a discrete time LTI system with impulse response
ℎ(𝑛) = 𝛽 𝑛 𝑢(−𝑛). Flip ℎ(𝑛) to get 𝛽 −𝑛 𝑢(𝑛). Shift by 𝑚 ≥ 0 to get
∞
1
𝑦(𝑚) = ∑ 𝛼 𝑛 𝑢(𝑛)𝛽−(𝑛−𝑚) 𝑢(𝑛 − 𝑚) = 𝛼 𝑚 𝛼
1−
𝑛=0 𝛽
𝛽𝑚
Similarly, for 𝑚 < 0, the output is 𝛼
1−
𝛽
Ans a
7. Given the signal 𝑥(𝑡) = 2 − |𝑡| for |𝑡| ≤ 1 and 0 otherwise, which is a triangular pulse,
input to an LTI system with impulse response ℎ(𝑡) = 𝑢(𝑡 + 3) − 𝑢(𝑡 − 3), which is a
rectangular pulse. To evaluate the convolution, we flip 𝑥(𝑡) and shift while evaluating
the area under the product. It can be seen that the peak is 3, which occurs when the
triangular pulse completely overlaps with the rectangular pulse. Further, this happens
for −2 ≤ 𝑡 ≤ 2
Ans d
8. This property is termed as Additivity
 Ans a
9. The even component of 𝑒 𝜎𝑡 , for 𝜎 real is given as cosh(𝜎𝑡)
Ans c
10. The input-output system corresponding to the ideal capacitor is causal but not
memoryless
Ans c
1. This property is termed as time-invariance
Ans c
𝑑𝑦(𝑡)
2. Given the system described by the linear differential equation 2 + 3𝑦(𝑡) = 𝑥(𝑡).
𝑑𝑡
Also 𝑦(0) = 2 and input signal 𝑥(𝑡) = 𝑒 −𝑡 . Let particular solution be 𝐶𝑒 −𝑡 . Then,
substituting in the DE
−2𝐶𝑒 −𝑡 + 3𝐶𝑒 −𝑡 = 𝑒 −𝑡 ⟹ 𝐶 = 1
Hence, particular solution is 𝑒 −𝑡 for 𝑡 > 0
Ans b
3. Given 𝑅 = 2Ω, 𝐿 = 4𝐻
As described in lectures, the input-output relation for this system is
𝐿 𝑑 𝑑
𝑖𝐿 (𝑡) + 𝑖𝐿 (𝑡) = 𝑖(𝑡) ⇒ 2 𝑖 (𝑡) + 𝑖𝐿 (𝑡) = 𝑖(𝑡)
𝑅 𝑑𝑡 𝑑𝑡 𝐿
Ans a
4. Given ℎ(𝑛) = 𝛼 𝑛 𝑢(−𝑛), 𝛼 > 1 and 𝑥(𝑛) = 𝑢(𝑛) − 𝑢(𝑛 − 𝑁 − 1)
1 − 𝛼 𝑛−1
𝛼 𝑛 𝑢(−𝑛) ∗ 𝑢(−𝑛) = (1 + 𝛼 −1 + ⋯ + 𝛼 𝑛 )𝑢(−𝑛) = 𝑢(−𝑛)
1 − 𝛼 −1
𝛼 𝑛 𝑢(−𝑛) ∗ (𝑢(𝑛) − 𝑢(𝑛 − 𝑁 − 1))
= 𝛼 𝑛 𝑢(−𝑛) ∗ (𝑢(𝑁 − 𝑛) − 𝑢(−1 − 𝑛))
= 𝛼 𝑛 𝑢(−𝑛) ∗ (𝑢(−(𝑛 − 𝑁)) − 𝑢(−(𝑛 + 1)))
1 − 𝛼 𝑛−𝑁−1 1 − 𝛼𝑛
= 𝑢(−𝑛 + 𝑁) − 𝑢(−𝑛 − 1)
1 − 𝛼 −1 1 − 𝛼 −1
For 𝑛 > 𝑁, this equals 0
1−𝛼𝑛−𝑁−1
For 0 ≤ 𝑛 ≤ 𝑁, this equals
1−𝛼−1
For 𝑛 ≤ −1, this equals
1 − 𝛼 𝑛−𝑁−1 1 − 𝛼 𝑛 𝛼 𝑛 − 𝛼 𝑛−𝑁−1 𝛼 𝑛 (1 − 𝛼 −(𝑁+1) )
− = =
1 − 𝛼 −1 1 − 𝛼 −1 1 − 𝛼 −1 1 − 𝛼 −1
Therefore, output is
0 for 𝑛>𝑁
1 − 𝛼 𝑛−𝑁−1
( ) for 0 ≤ 𝑛 ≤ 𝑁
= 1 − 𝛼 −1
𝛼 𝑛 (1 − 𝛼 −(𝑁+1) )
{ for 𝑛 ≤ −1
1 − 𝛼 −1
Ans a
5. The input-output relation for this system can be found as follows. Given 𝑣𝑐 (𝑡) is the
𝑑𝑣𝑐 (𝑡)
voltage across the capacitor. The current is 𝐶 , which implies that voltage across
𝑑𝑡
𝑑𝑣𝑐 (𝑡)
the resistance is 𝑅𝐶 . Therefore, using KVL, it follows that
𝑑𝑡
𝑑𝑣𝑐 (𝑡)
𝑣(𝑡) = 𝑣𝑐 (𝑡) + 𝑅𝐶
𝑑𝑡
Ans d
6. Given eigenfunction (𝑡) = 𝑒 𝛼𝑡 , for Re{𝛼} > −1. The output corresponding to impulse
response ℎ(𝑡) = 𝑒 −𝑡 𝑢(𝑡) is
∞ ∞
1
∫ 𝑒 −𝜏 𝑒 𝛼(𝑡−𝜏) 𝑑𝜏 = 𝑒 𝛼𝑡 ∫ 𝑒 −𝜏 𝑒 −𝛼𝜏 𝑑𝜏 = 𝑒 𝛼𝑡
0 0 1 + 𝛼
 1
1
The eigenvalue is 1+𝛼. The eigenvalue corresponding to eigenfunction 𝑒 2𝑡 is
1 2
=
1 3
1+2
Ans c
7. As shown in lectures, the impulse response of the system 𝛼𝑦(𝑛 − 1) + 𝑥(𝑛) is
𝛼 𝑛 𝑢(𝑛)
1
Hence, the system 2𝑦(𝑛) = 𝑦(𝑛 − 1) + 2𝑥(𝑛) ⇒ 𝑦(𝑛) = 2 𝑦(𝑛 − 1) + 𝑥(𝑛) has the
1 𝑛
impulse response (2) 𝑢(𝑛) = 2−𝑛 𝑢(𝑛)
Ans d
8. This property is termed as Homogeneity
Ans a
9. As shown in the lectures, the particular solution for this system is
𝐶
𝑒 −𝛽𝑡
𝛼−𝛽
Ans c
10. This can be simplified as follows
∞ ∞
1 𝑢
∫ sin(𝑢) 𝛿(𝛼𝑢 − 𝛽𝜋)𝑑𝑢 = ∫ sin ( ) 𝛿(𝑢 − 𝛽𝜋)𝑑𝑢
−∞ −∞ 𝛼 𝛼
∞
1 𝑢 + 𝛽𝜋 1 𝛽𝜋
=∫ sin ( ) 𝛿(𝑢)𝑑𝑢 = sin ( )
−∞ 𝛼 𝛼 𝛼 𝛼

Ans a
Assignment 4: Solution

1. Given the LTI system is stable, the ROC of 𝐻(𝑠) must include the 𝑗𝜔 axis
Ans a
2. Given signal
𝑥(𝑡) = 𝑒 −3𝑡 𝑢(𝑡 − 2) − 𝑒 5𝑡 𝑢(4 − 𝑡)
= 𝑒 −6 𝑒 −3(𝑡−2) 𝑢(𝑡 − 2) − 𝑒 20 𝑒 5(𝑡−4) 𝑢(−(𝑡 − 4))
𝑒 −6 𝑒 −2𝑠 𝑒 20 𝑒 −4𝑠
↔ +
𝑠+3 𝑠−5
ROC is −3 < 𝑠 < 5
Ans d
1
3. 𝑒 −𝑎𝑡 𝑢(−𝑡) has Laplace transform − 𝑠+𝑎
𝑑 1 1
𝑡𝑒 −𝑎𝑡 𝑢(−𝑡) has LT − 𝑑𝑠 (− 𝑠+𝑎) = − (𝑠+𝑎)2
𝑑 1 2
𝑡 × 𝑡𝑒 −𝑎𝑡 𝑢(−𝑡) has LT − (− (𝑠+𝑎)2) = − (𝑠+𝑎)3
𝑑𝑠
Ans d
1
4. Observe sin(𝜔0 𝑡) 𝑢(𝑡) = 2𝑗 (𝑒 𝑗𝜔0𝑡 − 𝑒 −𝑗𝜔0𝑡 )𝑢(𝑡). For 𝑅𝑒{𝑠} > 0, This has Laplace
transform
1 1 1 1 𝜔0
− = 2
2𝑗 𝑠 − 𝑗𝜔0 2𝑗 𝑠 + 𝑗𝜔0 𝑠 + 𝜔02
4
sin(−4𝑡) 𝑢(𝑡) = − sin(4𝑡) 𝑢(𝑡) ↔ −
𝑠 2 + 16
Ans a
5. The Laplace transform of 𝑡𝑥(−𝑡) can be found as follows
∞
𝑋(𝑠) = ∫ 𝑥(𝑡)𝑒 −𝑠𝑡 𝑑𝑡
−∞
∞ ∞
⇒ ∫ 𝑥(−𝑡)𝑒 −𝑠𝑡 𝑑𝑡 = ∫ 𝑥(𝑡)𝑒 𝑠𝑡 𝑑𝑡 = 𝑋(−𝑠)
−∞ −∞
∞
𝑑
⇒ ∫ −𝑡𝑥(−𝑡)𝑒 −𝑠𝑡 𝑑𝑡 = 𝑋(−𝑠)
−∞ 𝑑𝑠
∞
𝑑
⇒ ∫ 𝑡𝑥(−𝑡)𝑒 −𝑠𝑡 𝑑𝑡 = − 𝑋(−𝑠)
−∞ 𝑑𝑠

Ans c
6. Consider the signal 𝑥(𝑡) = 𝑒 −2𝑡 𝑢(𝑡) + 𝑒 4𝑡 𝑢(−𝑡).The Laplace transform 𝑋(𝑠) of this
signal is
1 1 6
− =−
𝑠+2 𝑠−4 (𝑠 + 2)(𝑠 − 4)
ROC is −2 < 𝑅𝑒{𝑠} < 4
Ans b
7. Give the Laplace transform of the impulse response of an LTI system is a polynomial
is 𝑠.
 𝑁 𝑁
𝑌(𝑠) 𝑘
𝑑𝑘 𝑥(𝑡)
= ∑ 𝑐𝑘 𝑠 ⟹ 𝑦(𝑡) = 𝑐0 𝑥(𝑡) + ∑ 𝑐𝑘
𝑋(𝑠) 𝑑𝑡𝑘
𝑘=0 𝑘=1
Hence, The relation between the input 𝑥(𝑡) and output 𝑦(𝑡) of the LTI system in the
time domain can be described by a linear constant coefficient differential equation
involving 𝑥(𝑡), 𝑦(𝑡) and higher order derivatives of 𝑥(𝑡) only
Ans c
𝑡
8. Given signal 𝑥(𝑡) has Laplace transform 𝑋(𝑠), then the Laplace transform of 𝑥 (𝛽) for
𝑡
𝛽 < 0 can be evaluated as follows. Set 𝛽 = 𝑢
∞ −∞ ∞
𝑡
∫ 𝑥 ( ) 𝑒 −𝑠𝑡 𝑑𝑡 = ∫ 𝛽𝑥(𝑢)𝑒 −𝑠𝛽𝑢 𝑑𝑡 = − ∫ 𝛽𝑥(𝑢)𝑒 −𝑠𝛽𝑢 𝑑𝑡 = −𝛽𝑋(𝛽𝑠)
−∞ 𝛽 ∞ −∞
𝑡 𝑡
Hence, 𝑥 (− 5) = 𝑥 (−5) has the LT −(−5)𝑋(−5𝑠) = 5𝑋(−5𝑠)
Ans b
9. Given signal
𝑥(𝑡) = 𝑒 −3𝑡 𝑢(𝑡 − 4) − 𝑒 −2𝑡 𝑢(2 − 𝑡)
= 𝑒 −12 𝑒 −3(𝑡−4) 𝑢(𝑡 − 4) − 𝑒 −4 𝑒 −2(𝑡−2) 𝑢(−(𝑡 − 2))
𝑒 −12 𝑒 −4𝑠 𝑒 −4 𝑒 −2𝑠
↔ +
𝑠+3 𝑠+2
ROC is −3 < 𝑠 < −2
Ans a
1
10. The inverse Laplace transform of 𝑠−𝛼 with ROC Re{𝑠} > 𝛼 is 𝑒 𝛼𝑡 𝑢(𝑡)
Ans b
 Solution-5
1
1. The coefficient of (𝑠−𝑝𝑖 )𝑟
is in the partial fraction expansion of 𝑋(𝑠) is
(𝑠 − 𝑝𝑖 )𝑟 𝑋(𝑠)|𝑠=𝑝𝑖
Ans a
−1
2. Given the Laplace transform 𝑋(𝑠) = (𝑠+1)(𝑠+2)2 with ROC Re{𝑠} > −1. Also, given
𝑐
1 𝜆
1 𝜆
2
the partial fraction expansion of 𝑋(𝑠) given as 𝑋(𝑠) = 𝑠+1 + 𝑠+2 + (𝑠+2) 2 . The value of

𝜆1 can be calculated as follows.

−1 1 1 −(𝑠 + 2)2 + (𝑠 + 2)(𝑠 + 1) + (𝑠 + 1)
+ + =
𝑠 + 1 𝑠 + 2 (𝑠 + 2)2 (𝑠 + 1)(𝑠 + 2)2
−1
=
(𝑠 + 1)(𝑠 + 2)2
−1 −1
𝑋(𝑠) = 2
⇒ (𝑠 + 2)2 𝑋(𝑠) =
(𝑠 + 1)(𝑠 + 2) (𝑠 + 1)
𝑑 1
⇒ (𝑠 + 2)2 𝑋(𝑠)| = | =1
𝑑𝑠 𝑠=−2 (𝑠 + 1)2 𝑠=−2
Ans c
𝑠+4
3. Consider 𝑠2 +5𝑠+6. This can be expressed as
2 1 𝑠+4
− = 2
𝑠 + 2 𝑠 + 3 𝑠 + 5𝑠 + 6
Since ROC is −3 < Re{𝑠} < −2, the corresponding inverse Laplace transform is
2 1
− ↔ −2𝑒 −2𝑡 𝑢(−𝑡) − 𝑒 −3𝑡 𝑢(𝑡)
𝑠+2 𝑠+3
(𝑠+4)𝑒 −3𝑠
Hence, inverse Laplace transform of 𝑋(𝑠) = 𝑠2 +5𝑠+6 is
−2𝑒 −2(𝑡−3) 𝑢(−𝑡 + 3) − 𝑒 −3(𝑡−3) 𝑢(𝑡 − 3)
= −2𝑒 6 𝑒 −2𝑡 𝑢(−𝑡 + 3) − 𝑒 9 𝑒 −3𝑡 𝑢(𝑡 − 3)
Ans c
4. Consider the signal 𝑥(𝑡) = 𝑒 −5𝑡 𝑢(−𝑡) + 𝑒 3𝑡 𝑢(𝑡). It can be seen that the ROC is
Re{𝑠} < −5, Re{𝑠} > 3
Hence, the Laplace transform does NOT exist
1 1 8
𝑒 −5𝑡 𝑢(−𝑡) + 𝑒 3𝑡 𝑢(𝑡) = − + =
𝑠 + 5 𝑠 − 3 (𝑠 + 5)(𝑠 − 3)
Ans b
𝑑2 𝑦(𝑡) 𝑑𝑦(𝑡)
5. Given DE 𝑑𝑡 2 − 2 𝑑𝑡 = 3𝑦(𝑡) + 𝑥(𝑡). Taking the Laplace transform
𝑌(𝑠) 1 1 1 1 1
= 𝐻(𝑠) = 2 = = ( − )
𝑋(𝑠) 𝑠 − 2𝑠 − 3 (𝑠 − 3)(𝑠 + 1) 4 𝑠 − 3 𝑠 + 1
Since given system is stable, ROC is −1 < Re{𝑠} < 3. Hence, the inverse Laplace
transform is
1 1
− 𝑒 3𝑡 𝑢(−𝑡) − 𝑒 −𝑡 𝑢(𝑡)
4 4

Ans a
 3 𝜔
6. Given Laplace transform of (𝑠+2)2 +9
= (𝑠+𝑎)20+𝜔2 with 𝜔0 = 3 and 𝑎 = 2, ROC
0
Re{𝑠} > −2 = −2. Hence, inverse Laplace transform is 𝑒 −𝑎𝑡 sin(𝜔0 𝑡) 𝑢(𝑡) =
𝑒 −2𝑡 sin(3𝑡) 𝑢(𝑡)
Ans b
𝑠2 +9𝑠+16
7. Given the Laplace transform 𝑋(𝑠) = (𝑠+2)(𝑠+3)2 with ROC Re{𝑠} > −2. Also, given
𝑐
1 𝜆
1 𝜆
2
the partial fraction expansion of 𝑋(𝑠) given as 𝑋(𝑠) = 𝑠+2 + 𝑠+3 + (𝑠+3) 2
. The value of
𝜆1 can be calculated as follows.
𝑠 2 + 9𝑠 + 16 2
𝑠 2 + 9𝑠 + 16
𝑋(𝑠) = ⟹ (𝑠 + 3) 𝑋(𝑠) =
(𝑠 + 2)(𝑠 + 3)2 (𝑠 + 2)
𝑠(𝑠 + 2) + 7(𝑠 + 2) + 2 2
= =𝑠+7+
(𝑠 + 2) 𝑠+2
𝑑 2
⟹ (𝑠 + 3)2 𝑋(𝑠)| =1− | = 1 − 2 = −1 = 𝜆1
𝑑𝑠 𝑠=−3 (𝑠 + 2)2 𝑠=−3
2 1 2 2(𝑠 + 3)2 − (𝑠 + 2)(𝑠 + 3) + 2(𝑠 + 2)
− + =
𝑠 + 2 𝑠 + 3 (𝑠 + 3)2 (𝑠 + 2)(𝑠 + 3)2
2
𝑠 + 9𝑠 + 16
=
(𝑠 + 2)(𝑠 + 3)2
Ans d
8. Given the LTI system is causal, the ROC of 𝐻(𝑠) is of the form Re{𝑠} > 𝜎max
Ans c
1
9. Observe − cos(4𝑡) 𝑢(−𝑡) = − 2 (𝑒 𝑗4𝑡 + 𝑒 −𝑗4𝑡 )𝑢(−𝑡). For Re{𝑠} < 0, this has
Laplace transform
1 1 1 1 𝑠
+ = 2
2 𝑠 − 4𝑗 2 𝑠 + 4𝑗 𝑠 + 16
Ans c
1
10. The inverse Laplace transform of 𝑠+𝛼, for ROC 𝑅𝑒{𝑠} > −𝛼, is 𝑒 −𝛼𝑡 𝑢(𝑡)
Ans a
 Solution-6

1. Given 𝑋(𝑧) = ∑∞
𝑛=−∞ 𝑥(𝑛)𝑧
−𝑛
. Hence,
∞ ∞ ∞
−𝑛 −(𝑛0 −𝑚) −𝑛0
∑ 𝑥(𝑛0 − 𝑛)𝑧 = ∑ 𝑥(𝑚)𝑧 =𝑧 ∑ 𝑥(𝑚)𝑧 𝑚
𝑛=−∞ 𝑚=−∞ 𝑚=−∞
∞
1
= 𝑧 −𝑛0 ∑ 𝑥(𝑚)(𝑧 −1 )−𝑚 = 𝑧 −𝑛0 𝑋 ( )
𝑧
𝑚=−∞

Ans c
2. The z-transform of the impulse response corresponding to an LTI system described by
a difference equation is a Rational function of 𝑧
Ans c
1
3. Given signal 5−𝑛+1 𝑢(𝑛 + 1) − 4𝑛 𝑢(−𝑛 − 1). ROC of 5−𝑛+1 𝑢(𝑛 + 1) is |𝑧| > 5 and
ROC of 4𝑛 𝑢(−𝑛 − 1) is |𝑧| < 4.
1
Hence, ROC is < |𝑧| < 4
5
Ans b
𝜋
4. The 𝑧 −transform of cos(Ω0 𝑛) 𝑢(𝑛) = cos (3 𝑛) 𝑢(𝑛) is
𝑧
𝑧 2 − 𝑧 cos Ω0 𝑧2 − 2
=
𝑧 2 − 2𝑧 cos Ω0 + 1 𝑧 2 − 𝑧 + 1
Ans a
5. Given 𝑋(𝑧) = ∑∞𝑛=−∞ 𝑥(𝑛)𝑧
−𝑛
. Hence,
∞ ∞
𝑑 𝑑
𝑋(𝑧) = ∑ −𝑛𝑥(𝑛)𝑧 −𝑛−1 ⇒ −𝑧 𝑋(𝑧) = ∑ 𝑛𝑥(𝑛)𝑧 −𝑛
𝑑𝑧 𝑑𝑧
𝑛=−∞ 𝑛=−∞
2 ∞
𝑑 𝑑 𝑑 𝑑
⇒ (−𝑧 𝑋(𝑧)) = − 𝑋(𝑧) − 𝑧 2 𝑋(𝑧) ↔ ∑ −𝑛2 𝑥(𝑛)𝑧 −𝑛−1
𝑑𝑧 𝑑𝑧 𝑑𝑧 𝑑𝑧
𝑛=−∞
2 ∞
𝑑 𝑑
⇒ 𝑋(𝑧) + 𝑧 2 𝑋(𝑧) ↔ ∑ 𝑛2 𝑥(𝑛)𝑧 −𝑛−1
𝑑𝑧 𝑑𝑧
𝑛=−∞
2 ∞
𝑑 𝑑
⇒𝑧 𝑋(𝑧) + 𝑧 2 2 𝑋(𝑧) ↔ ∑ 𝑛2 𝑥(𝑛)𝑧 −𝑛
𝑑𝑧 𝑑𝑧
𝑛=−∞

Ans d
6. We have 𝑋(𝑧) = ∑∞
𝑛=−∞ 𝑥(𝑛)𝑧
−𝑛
. Since 𝑥(𝑛) is real,
∞ ∞
∗ (𝑧) ∗ (𝑛)(𝑧 ∗ )−𝑛
𝑋 = ∑ 𝑥 = ∑ 𝑥(𝑛)(𝑧 ∗ )−𝑛 = 𝑋(𝑧 ∗ )
𝑛=−∞ 𝑛=−∞
Hence, if 𝑧0 is a zero of 𝑋(𝑧), then 𝑧0∗ is also a zero of 𝑋(𝑧)
Ans a
7. Given 𝑋(𝑧) = ∑∞ 𝑛=−∞ 𝑥(𝑛)𝑧
−𝑛
. Hence,
 ∞ ∞
𝑑 𝑑
𝑋(𝑧) = ∑ −𝑛𝑥(𝑛)𝑧 −𝑛−1 ⇒ −𝑧 𝑋(𝑧) = ∑ 𝑛𝑥(𝑛)𝑧 −𝑛
𝑑𝑧 𝑑𝑧
𝑛=−∞ 𝑛=−∞

Ans c
𝑑𝑋(𝑧)
8. Note that 𝑛𝑥(𝑛) has 𝑧 −transform −𝑧 . Hence, 𝑛𝑎𝑛 𝑢[𝑛] has 𝑧 −transform
𝑑𝑧
𝑑 𝑧 𝑑 𝑎 𝑎 𝑎𝑧 𝑎𝑧 −1
−𝑧 = −𝑧 (1 + ) = −𝑧 × − = =
𝑑𝑧 𝑧 − 𝑎 𝑑𝑧 𝑧−𝑎 (𝑧 − 𝑎)2 (𝑧 − 𝑎)2 (1 − 𝑎𝑧 −1 )2
Ans b
1 −𝑛+1 1 𝑛+2
9. Given signal (2) 𝑢(−𝑛 − 4) + (6) 𝑢(𝑛 − 2). Its ROC can be evaluated as
follows
1 −𝑛+1 1
( ) 𝑢(−𝑛 − 4) = × 2𝑛 𝑢(−𝑛 − 4) ⇒ |𝑧| < 2
2 2
1 𝑛+2 1 2 1 𝑛 1
( ) 𝑢(𝑛 − 2) = ( ) × ( ) 𝑢(𝑛 − 2) ⇒ |𝑧| >
6 6 6 6
1
ROC is 6 < |𝑧| < 2
Ans a
10. For a discrete time LTI system to be neither causal nor anti-causal, the ROC of the
𝑧 −transform of its impulse response must be of the form 𝑟1 < |𝑧| < 𝑟2
Ans b
 Solution-7
1. For a discrete time LTI system to be anti-causal, the ROC of the z-transform of its
impulse response must be of the form |𝑧| < 𝑟min
Ans d
2𝑧 2 −11𝑧 3𝑧 𝑧
2. It can be seen that = 𝑧+2 − 𝑧−3. Give system is causal, ROC is |𝑧| > 3, the
𝑧 2 −𝑧−6
inverse z-transform is
3 × (−2)𝑛 𝑢(𝑛) − 3𝑛 𝑢(𝑛)
Ans a
𝑧3
3. The inverse z-transform of 𝑧−𝑎 with ROC |𝑧| < |𝑎| can be found as follows
𝑧
↔ −𝑎𝑛 𝑢(−𝑛 − 1)
𝑧−𝑎
𝑧2 𝑧
⇒ =𝑧 ↔ −𝑎𝑛+1 𝑢(−𝑛 − 2)
𝑧−𝑎 𝑧−𝑎
𝑧3 𝑧2
⇒ =𝑧 ↔ −𝑎𝑛+2 𝑢(−𝑛 − 3)
𝑧−𝑎 𝑧−𝑎
Ans c
−1
4. The inverse 𝑧 −transform of 𝑒 𝑎𝑧 with ROC |𝑧| > |𝑎| can be evaluated as follows.
−1 1 1 1
𝑒 𝑎𝑧 = 1 + 𝑎𝑧 −1 + 𝑎2 𝑧 −2 + 𝑎3 𝑧 −3 + ⋯ ↔ 𝑎𝑛 𝑢(𝑛)
2 3! 𝑛!
Ans c
5. For a discrete time LTI system to be BIBO stable, the ROC of the 𝑧 −transform of its
impulse response must include the unit-circle
Ans a
2𝑧 2 −5𝑧 𝑧 𝑧
6. It can be seen that = 𝑧−3 + 𝑧−2. Given ROC is 2 < |𝑧| < 3, the inverse
𝑧 2 −5𝑧+6
𝑧 −transform is
−3𝑛 𝑢(−𝑛 − 1) + 2𝑛 𝑢(𝑛)
Ans d
sin(Ω )𝑧
7. The 𝑧 −transform of sin(Ω𝑜 𝑛) 𝑢(𝑛) is 𝑧 2−(2 cos 𝑜Ω
𝑜 )𝑧+1
𝜋
𝑧 (sin )𝑧
4
Given z-transform of 𝑧 2−√2𝑧+1 = √2 𝜋 with ROC |𝑧| > 1. Hence the inverse
𝑧 2 −2 cos 𝑧+1
4
𝜋
𝑧 −transform is √2 sin ( 4 𝑛) 𝑢(𝑛)
Ans c
𝑧
8. The inverse z-transform of , with ROC |𝑧| < |𝑎|, is −𝑎𝑛 𝑢(−𝑛 − 1)
𝑧−𝑎
𝑧
Hence, inverse z-transform of − 𝑧+𝑎, with ROC |𝑧| < |𝑎|, is (−𝑎)𝑛 𝑢(−𝑛 − 1)
Ans b
1
9. The coefficient 𝜆𝑟−𝑘 corresponding to the term (𝑧−𝑝𝑖 )𝑟−𝑘
in the partial fraction
𝑋(𝑧) 1 𝑑𝑘 𝑋(𝑧)
expansion of is 𝑘! 𝑑𝑧 𝑘 (𝑧 − 𝑝𝑖 )𝑟 |
𝑧 𝑧 𝑧=𝑝𝑖
1 1
Now set 𝑘 = 𝑟 − 1. Hence, 𝜆1 corresponding to (𝑧−𝑝 )𝑟−(𝑟−1) = 𝑧−𝑝 is
𝑖 𝑖
1 𝑑 𝑟−1 𝑟
𝑋(𝑧)
(𝑧 − 𝑝𝑖 ) |
(𝑟 − 1)! 𝑑𝑧 𝑟−1 𝑧 𝑧=𝑝
𝑖
 Ans a
3𝑧 2 −19𝑧 4𝑧 𝑧
10. It can be seen that = 𝑧+3 − 𝑧−4. Give system is stable, ROC is |𝑧| < 3, the
𝑧 2 −𝑧−12
inverse z-transform is
−4 × (−3)𝑛 𝑢(−𝑛 − 1) + 4𝑛 𝑢(−𝑛 − 1)
Ans a
 Solution-8
1. The condition 𝑥(𝑡) can be unbounded at a finite number of discontinuities in any
interval is NOT one of the Dirichlet conditions. In fact, the Dirichlet conditions state
that 𝑥(𝑡) is finite at all the discontinuities
Ans b
1
2. The Fourier transform of the unit-step function is 𝜋𝛿(𝜔) + 𝑗𝜔
Ans d
3. Given periodic signal 𝑥(𝑡) with fundamental frequency 𝜔0 and complex exponential
Fourier series ∑∞𝑘=−∞ 𝑐𝑘 𝑒
𝑗𝑘𝜔0 𝑡
, its power is ∑∞ 2
𝑘=−∞|𝑐𝑘 | . This follows from the
Parseval’s theorem
Ans a
4. Given
∞
𝑋(𝜔) = ∫ 𝑥(𝑡)𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
∞
⇒ 𝑋(𝑡) = ∫ 𝑥(𝜔)𝑒 −𝑗𝜔𝑡 𝑑𝜔
−∞
1 1 ∞ 1 −𝑗𝜔𝑡
1 ∞ 1
⇒ 2 𝑋(𝑡) = ∫ 𝑥(𝜔)𝑒 𝑑𝜔 = ∫ 𝑥(−𝜔)𝑒 𝑗𝜔𝑡 𝑑𝜔
4𝜋 2𝜋 −∞ 2𝜋 2𝜋 −∞ 2𝜋
𝑋(𝑡) 1
Hence, ↔ 2𝜋 𝑥(−𝜔)
4𝜋 2
Ans c
𝑑
5. The Fourier transform of 𝑥(𝑡) is 𝑗𝜔𝑋(𝜔)
𝑑𝑡
Ans a
𝜋
6. The CEFS coefficients of cos2 (𝑡 + 6 ) can be found as follows
𝜋 1 𝜋 1 1 𝑗𝜋 𝑗𝜋
cos2 (𝑡 + ) = (1 + cos (2𝑡 + )) = + (𝑒 𝑗2𝑡 𝑒 3 + 𝑒 −𝑗2𝑡 𝑒 − 3 )
6 2 3 2 4
1 1 1 √3 1 1 √3
Therefore, the CEFS coefficients are 2 , 4 (2 + 𝑗 2 ) , 4 (2 − 𝑗 2 )
Ans c
7. The Fourier transform of the signal function 𝑥(𝑡) = −sgn(𝑡) can be evaluated as
follows
𝑑𝑥(𝑡)
= −2𝛿(𝑡) ↔ −2 = 𝑗2𝜋𝑓𝑋(𝑓)
𝑑𝑡
−2 2𝑗
⇒ 𝑋(𝑓) = =
𝑗2𝜋𝑓 𝜔
Ans a
8. Given a real periodic signal 𝑥(𝑡) with fundamental frequency 𝜔0 .
∞ ∞
𝑗𝑘𝜔0 𝑡
𝑥(𝑡) = ∑ 𝑐𝑘 𝑒 = 𝑎0 + ∑(𝑎𝑘 cos(𝑘𝜔0 𝑡) + 𝑏𝑘 sin(𝑘𝜔0 𝑡))
𝑘=−∞ 𝑘=1
The coefficient 𝑐𝑘 corresponding to 𝑒 𝑗𝑘𝜔0𝑡 in the complex exponential Fourier series
and 𝑎𝑘 corresponding to cos(𝑘𝜔0 𝑡) in the trigonometric Fourier series are related as
𝑎𝑘 = 𝑐𝑘 + 𝑐−𝑘
Since signal is real, we must have
 ∞

𝑥 ∗ (𝑡)
= ∑ 𝑐𝑘∗ 𝑒 −𝑗𝑘𝜔0𝑡 = 𝑥(𝑡) ⇒ 𝑐−𝑘 = 𝑐𝑘∗
𝑘=−∞
Therefore,
𝑎𝑘 = 𝑐𝑘 + 𝑐−𝑘 = 𝑐𝑘 + 𝑐𝑘∗ = 2 Re{𝑐𝑘 }
Ans d
𝑡2 𝜔2
9. The FT of the Gaussian pulse 𝑒 − 𝑎 is 𝑋(𝜔) = √𝜋𝑎𝑒 −𝑎 4 . Hence,
𝑡2 𝜔2
− −𝑎
𝑒 2𝑎 ↔ √𝜋2𝑎𝑒 2
32
𝑋(0) = 8 ⟹ √2𝜋𝑎 = 8 ⟹ 2𝜋𝑎 = 64 ⇒ 𝑎 =
𝜋
Ans d
∞
10. The quantity ∫−∞ 𝑋(𝜔)𝑌 ∗ (𝜔
̃ − 𝜔)𝑑𝜔 represents the convolution of 𝑋(𝜔), 𝑌 ∗ (𝜔). It is
∞
the FT of 2𝜋𝑥(𝑡)𝑦 ∗ (−𝑡). Further, ∫−∞ 𝑋(𝜔)𝑌 ∗ (−𝜔)𝑑𝜔 is the convolution evaluated
̃ = 0. Hence, it equals the FT of 2𝜋𝑥(𝑡)𝑦 ∗ (−𝑡) at 𝜔 = 0, which is
for 𝜔
∞ ∞
2𝜋 ∫ 𝑥(𝑡)𝑦 ∗ (−𝑡)𝑒 −𝑗𝜔𝑡 𝑑𝑡| = 2𝜋 ∫ 𝑥(𝑡)𝑦 ∗ (−𝑡)𝑑𝑡
−∞ 𝜔=0 −∞
Ans a
 Solution-9
1
1. The impulse response of the Hilbert transform is 𝜋𝑡
Ans a
2. The Parseval’s relation for a continuous time signal 𝑥(𝑡) is given as
∞
1 ∞
∫ |𝑥(𝑡)|2 𝑑𝑡 = ∫ |𝑋(𝜔)|2 𝑑𝜔
−∞ 2𝜋 −∞
Ans d
(100+𝑗𝜔)
3. Given the transfer function 𝐻(𝜔) = 𝐻(𝜔) = (1+𝑗𝜔)(10+𝑗𝜔)(1000+𝑗𝜔)2. For 𝜔 ≫ 1000
this can be approximated as
𝑗𝜔 1
𝐻(𝜔) = =
𝑗𝜔 × 𝑗𝜔 × (𝑗𝜔)2 𝑗𝜔
Hence the Bode plot of |𝐻(𝜔)| decreases as 60 dB/ decade
Ans c
1
4. The 3 𝑑𝐵 frequency of the 𝑅𝐶 filter in rad/s is
𝑅𝐶
Ans c
(100+𝑗𝜔)2
5. Given frequency response is 𝐻(𝜔) = (1+𝑗𝜔)(10+𝑗𝜔)(1000+𝑗𝜔). Between 100 ≪ 𝜔 ≪
1000 the Bode magnitude plot is approximately
(100 + 𝑗𝜔)2 (𝑗𝜔)2
≈
(1 + 𝑗𝜔)(10 + 𝑗𝜔)(1000 + 𝑗𝜔) 𝑗𝜔 × 𝑗𝜔 × 1000
Hence, it is constant
Ans b
1
6. Given the signal cos(2𝜋60𝑡), its FT is 2 (𝛿(𝑓 − 60) + 𝛿(𝑓 + 60)). After sampling by
the unit impulse train with sampling frequency 𝑓𝑠 = 150 𝐻𝑧, resulting spectrum is
∞
𝑓𝑠 𝑓𝑠
𝑓𝑠 ∑ 𝑋(𝑓 − 𝑘𝑓𝑠 ) = (𝛿(𝑓 − 60) + 𝛿(𝑓 + 60)) + (𝛿(𝑓 − 210) + 𝛿(𝑓 − 90))
2 2
𝑘=−∞
𝑓𝑠
+ (𝛿(𝑓 + 90) + 𝛿(𝑓 + 210)) + ⋯
2
Hence one can apply an ideal LPF with any cut-off frequency 60 𝐻𝑧 < 𝑓𝑐 < 90 𝐻𝑧
Ans a
1
7. The Fourier transform of 𝑥(𝑡) = 𝑒 −𝑎𝑡 , 𝑎 > 0 is 𝑗𝜔+𝑎. Hence, the Fourier transform of
𝑑 1 1 2 1 2
𝑡𝑒 −𝑎𝑡 is 𝑗 𝑑𝜔 𝑗𝜔+𝑎 = 𝑗 × −𝑗 (𝑗𝜔+𝑎) = (𝑗𝜔+𝑎)
Ans c
8. The coefficients are given as
𝑇0 𝑇0
1 2 1 2
∫ 𝑥(𝑡)𝑒 −𝑗𝑘𝜔0𝑡 𝑑𝑡 = × 𝑇0 ∫ 𝛿(𝑡)𝑒 −𝑗𝑘𝜔0𝑡 𝑑𝑡 = 1
𝑇0 −𝑇0 𝑇0 𝑇
− 0
2 2
Ans a
9. The Fourier transform of signal 𝑥(𝑡) = 𝑒 −𝑎|𝑡| , 𝑎 > 0 is
 1 1 2𝑎
𝑒 −𝑎𝑡 𝑢(𝑡) + 𝑒 𝑎𝑡 𝑢(−𝑡) = − = 2
𝑗𝜔 + 𝑎 𝑗𝜔 − 𝑎 𝜔 + 𝑎2
Ans c
10. As shown in lecture, for a real periodic signal, 𝑎𝑘 = 2 Re{𝑐𝑘 }
Ans b
 Solution-10
𝜔2 𝜔2
2 𝜋 𝜋
1. The FT of the Gaussian pulse 𝑒 −𝑎𝑡 is 𝑋(𝜔) = √𝑎 𝑒 − 4𝑎 . Hence, √ 4 𝑒 − 8 has the
inverse FT
𝜋 𝜔2 1 𝜋 −𝜔 2 1 −2𝑡 2
√ 𝑒− 8 = √ 𝑒 8 ↔ 𝑒
4 √2 2 √2
Ans c
2. Given 𝑥(𝑡) = cos(2𝜋𝑓0 𝑡) ⟹ 𝑥̂(𝑡) = sin(2𝜋𝑓0 𝑡)
𝑥(𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑥̂(𝑡) sin(2𝜋𝑓𝑐 𝑡)
= cos(2𝜋𝑓0 𝑡) cos(2𝜋𝑓𝑐 𝑡) − sin(2𝜋𝑓0 𝑡) sin(2𝜋𝑓𝑐 𝑡)
= cos(2𝜋(𝑓𝑐 + 𝑓0 )𝑡)
Ans d
1
3. Given the signal cos(2𝜋100𝑡), its FT is 2 (𝛿(𝑓 − 100) + 𝛿(𝑓 + 100)). After sampling
by the unit impulse train with sampling frequency 𝑓𝑠 = 250 𝐻𝑧, resulting spectrum is
∞

𝑓𝑠 ∑ 𝑋(𝑓 − 𝑘𝑓𝑠 )
𝑘=−∞
𝑓𝑠 𝑓𝑠
= (𝛿(𝑓 − 100) + 𝛿(𝑓 + 100)) + (𝛿(𝑓 − 350) + 𝛿(𝑓 − 150))
2 2
𝑓𝑠
+ (𝛿(𝑓 + 150) + 𝛿(𝑓 + 350)) + ⋯
2
Hence one can apply an ideal LPF with any cut-off frequency 100 𝐻𝑧 < 𝑓𝑐 < 150 𝐻𝑧
Ans a
2𝜋 2𝜋
4. As shown in the lectures, 𝑇𝑠 = 𝜔 , when sampled by ∑∞
𝑘=−∞ 𝛿 (𝑡 − 𝑘 𝜔 ) the Fourier
𝑠 𝑠
transform of the resulting signal is
∞ ∞
1 𝜔𝑠
∑ 𝑋(𝜔 − 𝑘𝜔𝑠 ) = ∑ 𝑋(𝜔 − 𝑘𝜔𝑠 )
𝑇𝑠 2𝜋
𝑘=−∞ 𝑘=−∞
𝜔𝑠 ∞ 2𝜋
Hence, when sampled by ∑ 𝛿 (𝑡 − 𝑘 𝜔 ), FT is
2𝜋 𝑘=−∞ 𝑠
∞
𝜔𝑠2
∑ 𝑋(𝜔 − 𝑘𝜔𝑠 )
4𝜋 2
𝑘=−∞
Ans c
5. Given 𝑥(𝑡) = sin(2𝜋𝑓0 𝑡) ⟹ 𝑥̂(𝑡) = − cos(2𝜋𝑓0 𝑡)
𝑥(𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑥̂(𝑡) sin(2𝜋𝑓𝑐 𝑡)
= sin(2𝜋𝑓0 𝑡) cos(2𝜋𝑓𝑐 𝑡) + cos(2𝜋𝑓0 𝑡) sin(2𝜋𝑓𝑐 𝑡)
= sin(2𝜋(𝑓𝑐 + 𝑓0 )𝑡)
Ans b
sin(𝑎𝑡)
6. The Fourier transform of is 1 for |𝜔| ≤ 𝑎 and 0 otherwise
𝜋𝑡
sin(2𝑡)
Hence, Fourier transform of is 𝜋 for |𝜔| ≤ 2 and 0 otherwise
𝑡
 2
sin(2𝑡) 2 sin (𝜋 𝜋 𝑡) 1
=𝜋 ↔ 𝜋 for |𝑓| ≤ , 0 otherwise
𝑡 𝜋 𝜋2𝑡 𝜋
𝜋
= 𝜋 for |𝜔| ≤ 2, 0 otherwise
Ans a
7. Given 𝑅𝐶 circuit with 𝑅 = 36 𝑀Ω and 𝐶 = 285 𝑝𝐹. The rise time is given as
2.1972 𝑅𝐶 = 2.1972 × 36 × 106 × 285 × 10−12 ≈ 22.5 𝑚𝑠
Ans c
cos(2𝜔) sin(𝜔)
8. Given signal 𝑥(𝑡) has the Fourier transform 𝑋(𝜔) = 2 . Use the principle
𝜔
sin(𝑎𝜔) sin(𝜔) 1
2 has IFT 𝑝2𝑎 (𝑡) = 1 for |𝑡| ≤ 𝑎 and 0 otherwise. Hence, has IFT 2 𝑝2 (𝑡).
𝜔 𝜔
1 1 cos(2𝜔) sin(𝜔)
Further, cos 2𝜔 has IFT 2 𝛿(𝑡 + 2) + 2 𝛿(𝑡 − 2). Therefore, 2 has IFT
𝜔
1 1 1
𝑝 (𝑡 − 2) + 2 𝑝2 (𝑡 + 2) = 𝑥(𝑡). It follows that 𝑥(−2) = 2
2 2
sin(𝜔) sin(2𝜋𝑓) 1 sin(𝜋𝑓2) 1
= = ×2 ↔ 𝑝2 (𝑡)
𝜔 2𝜋𝑓 2 𝜋𝑓2 2
1 1
cos(2𝜔) = cos(4𝜋𝑓) = cos(2𝜋2𝑓) ↔ 𝛿(𝑡 − 2) + 𝛿(𝑡 + 2)
2 2
sin(𝜔) 1 1 1
2 cos(2𝜔) × ↔ 2 × 𝑝2 (𝑡) ∗ ( 𝛿(𝑡 − 2) + 𝛿(𝑡 + 2))
𝜔 2 2 2
1 1 1
= 𝑝2 (𝑡 − 2) + 𝑝2 (𝑡 + 2) ⇒ 𝑥(−2) =
2 2 2
Ans a
sin(8𝜋𝑡) sin(8𝜋𝑡) 1
9. Given cos(8𝜋𝑡). has Fourier transform 𝑝 (𝑓). Hence,
2𝜋𝑡 2𝜋𝑡 2 8
sin(8𝜋𝑡) 1 1 1
cos(8𝜋𝑡) has Fourier transform 4 𝑝8 (𝑓 − 4) + 4 𝑝8 (𝑓 + 4) = 4 𝑝16 (𝑓), where
2𝜋𝑡
𝑝16 (𝑓) = 1 for −8 ≤ 𝑓 ≤ 8. Hence, the minimum sampling frequency required is
twice the maximum frequency = 16 𝐻𝑧
Ans d
1
10. Given the signal cos(10000𝜋𝑡), its FT is 2 (𝛿(𝑓 − 5000) + 𝛿(𝑓 + 5000)). After
sampling by the unit impulse train with sampling frequency 𝑓𝑠 = 7.5 𝑘𝐻𝑧, resulting
spectrum is
∞
𝑓𝑠
𝑓𝑠 ∑ 𝑋(𝑓 − 𝑘𝑓𝑠 ) = (𝛿(𝑓 − 5𝑘) + 𝛿(𝑓 + 5𝑘))
2
𝑘=−∞
𝑓𝑠 𝑓𝑠
+ (𝛿(𝑓 − 12.5𝑘) + 𝛿(𝑓 − 2.5𝑘)) + (𝛿(𝑓 + 2.5𝑘) + 𝛿(𝑓 + 12.5𝑘))
2 2
𝑓𝑠 𝑓𝑠
+ (𝛿(𝑓 − 20𝑘) + 𝛿(𝑓 − 10𝑘)) + (𝛿(𝑓 + 10𝑘) + 𝛿(𝑓 + 20𝑘)) + ⋯
2 2
𝑓𝑠 𝑓𝑠
+ (𝛿(𝑓 − 27.5𝑘) + 𝛿(𝑓 − 17.5𝑘)) + (𝛿(𝑓 + 27.5𝑘) + 𝛿(𝑓 + 17.5𝑘)) + ⋯
2 2

After ideal LPF with cutoff frequency 𝑓𝑐 = 11 𝑘𝐻𝑧, output spectrum is

𝑓𝑠 𝑓𝑠
(𝛿(𝑓 − 5𝑘) + 𝛿(𝑓 + 5𝑘)) + (𝛿(𝑓 + 2.5𝑘) + 𝛿(𝑓 − 2.5𝑘))
2 2
 𝑓𝑠
+ (𝛿(𝑓 + 10𝑘) + 𝛿(𝑓 − 10𝑘)) + ⋯
2
Output spectrum contains sinusoids 2.5 𝑘𝐻𝑧, 5 𝑘𝐻𝑧, 10 𝑘𝐻𝑧 only
Ans c
 Solution-11
1. Consider 𝑥(𝑛) has DTFT 𝑋(Ω).
∞

𝑋(Ω) = ∑ 𝑥(𝑛)𝑒 −𝑗Ω𝑛

𝑛=−∞
∞ ∞
𝑑𝑋(Ω) 𝑑𝑋(Ω)
= ∑ −𝑗𝑛𝑥(𝑛)𝑒 −𝑗Ω𝑛 ⇒ 𝑗 = ∑ 𝑛𝑥(𝑛)𝑒 −𝑗Ω𝑛
𝑑Ω 𝑑Ω
𝑛=−∞ 𝑛=−∞
2 ∞ 2 ∞
𝑑 𝑋(Ω) 2 −𝑗Ω𝑛
𝑑 𝑋(Ω)
⇒𝑗 = ∑ −𝑗𝑛 𝑥(𝑛)𝑒 ⇒ − = ∑ 𝑛2 𝑥(𝑛)𝑒 −𝑗Ω𝑛
𝑑Ω2 𝑑Ω2
𝑛=−∞ 𝑛=−∞
2
Therefore, 𝑛 𝑥(𝑛) has DTFT
𝑑 2 𝑋(Ω)
−
𝑑Ω2
Ans d
2. Given 𝑋(Ω) = 1, |Ω| ≤ 𝑊 = 2 and 0 for 𝑊 = 2 < |Ω| ≤ 𝜋. As shown in lectures, its
inverse DTFT is
sin(𝑊𝑛) sin(2𝑛)
=
𝜋𝑛 𝜋𝑛
Ans b
1 1
3. The DTFT of – sin(Ω0 𝑛) = − 2𝑗 𝑒 𝑗Ω0𝑛 + 2𝑗 𝑒 −𝑗Ω0 𝑛 is
𝑗𝜋𝛿(Ω − Ω0 ) − 𝑗𝜋𝛿(Ω + Ω0 )
Ans d
𝑁−1 𝑁+1
4. Given 𝑥(𝑛) = 𝑢 (𝑛 + ) − 𝑢 (𝑛 − ), where 𝑢(𝑛) is the unit-step signal.
2 2
(𝑁−1) Ω𝑁
−𝑗Ω sin( )
2
Note 𝑦(𝑛) = 𝑢(𝑛) − 𝑢(𝑛 − 𝑁) has the DTFT 𝑒 2
Ω
sin( )
2
𝑁−1 𝑁−1 𝑁+1
The DTFT of 𝑥(𝑛) = 𝑦 (𝑛 + ) = 𝑢 (𝑛 + ) − 𝑢 (𝑛 − 2 ) is
2 2
Ω𝑁 Ω𝑁
𝑁−1 (𝑁−1) sin ( ) sin ( 2 )
𝑗Ω
𝑒 2 𝑒 −𝑗Ω
2
2 =
Ω Ω
sin ( 2 ) sin ( 2 )
Ans c
5. The IDFT coefficients 𝑥(𝑛) of the finite length sequence 𝑋(𝑘), 0 ≤ 𝑘 ≤ 𝑁 − 1, are
𝑘𝑛
1 𝑗2𝜋
defined as 𝑥(𝑛) = 𝑁 ∑𝑁−1
𝑘=0 𝑋(𝑘)𝑒 𝑁

Ans b
6. Discrete Time Fourier Transform (DTFT) is best suited to represent a discrete time
aperiodic signal 𝑥(𝑛) of infinite duration
Ans a
7. 𝑦(𝑛) = ∑𝑛𝑘=−∞ 𝑥(𝑘) = 𝑥(𝑘) ∗ 𝑢(𝑘). Hence,
1 𝑋(Ω)
𝑌(Ω) = 𝑋(Ω)𝑈(Ω) = 𝑋(Ω) (𝜋𝛿(Ω) + −𝑗Ω
) = 𝜋𝑋(0)𝛿(Ω) +
1−𝑒 1 − 𝑒 −𝑗Ω
Ans d
8. The DTFT of cos(Ω0 𝑛) is
𝜋𝛿(Ω − Ω0 ) + 𝜋𝛿(Ω + Ω0 )
Ans c
9. Given 𝑋(Ω) = 1, |Ω| ≤ 𝑊 and 0 for 𝑊 < |Ω| ≤ 𝜋. As shown in lectures, its inverse
DTFT is
sin(𝑊𝑛)
𝜋𝑛
Ans d
2𝜋
10. The 2 × 2 DFT matrix is given as follows. 𝑊2 = 𝑒 −𝑗 2 = 𝑒 −𝑗𝜋 = −1. The DFT matrix
is
1 1 1 1
[ ]=[ ]
1 𝑊2−1 1 −1
Ans c
 Solution-12
1. Given phase response 𝜙(𝜔). Its group and phase delays, respectively, at 𝜔 = 𝜔0 are
given as
𝑑𝜙(𝜔) 𝜙(𝜔0 )
− | ,−
𝑑𝜔 𝜔=𝜔 𝜔0
0

Ans c
𝑒 −𝑗Ω
2. The IDFT of 2 can be evaluated as follows
(1−𝑎𝑒 −𝑗Ω )
1
𝑎𝑛 𝑢(𝑛) ↔
1 − 𝑎𝑒 −𝑗Ω
𝑒 −𝑗Ω
𝑎𝑛−1 𝑢(𝑛 − 1) ↔
1 − 𝑎𝑒 −𝑗Ω
2
𝑛 𝑛−1 −𝑗Ω
1
⇒ 𝑎 𝑢(𝑛) ∗ 𝑎 𝑢(𝑛 − 1) ↔ 𝑒 ( )
1 − 𝑎𝑒 −𝑗Ω
Observe that
∞
𝑎𝑛 𝑢(𝑛) ∗ 𝑎𝑛−1 𝑢(𝑛 − 1) = ∑ 𝑎𝑚 𝑢(𝑚)𝑎𝑛−1−𝑚 𝑢(𝑛 − 1 − 𝑚)
𝑚=−∞
𝑛−1
= (∑ 𝑎𝑛−1 ) 𝑢(𝑛 − 1) = 𝑛𝑎𝑛−1 𝑢(𝑛 − 1)
𝑚=0
This can also be seen as follows
2
1
( ) ↔ (𝑛 + 1)𝑎𝑛 𝑢(𝑛)
1 − 𝑎𝑒 −𝑗Ω
2
1
⇒( ) 𝑒 −𝑗Ω ↔ 𝑛𝑎𝑛−1 𝑢(𝑛 − 1)
1 − 𝑎𝑒 −𝑗Ω
Ans a
3. The DFT is given as
{2,3,2,3,2,3,2,3}
3 3
2𝑛𝑘 𝑘 2𝑛𝑘
𝑗2𝜋 𝑗2𝜋
𝑋(𝑘) = 2 ∑ 𝑒 8 + 3𝑒 8 ∑ 𝑒 𝑗2𝜋 8
𝑛=0 𝑛=0
3 3
𝑗𝜋𝑘 𝜋𝑘 𝑗𝜋𝑘
𝑛
= 2∑𝑒 2 + 3𝑒 𝑗 4 ∑ 𝑒 2
𝑛

𝑛=0 𝑛=0
3
𝜋𝑘 𝑗𝜋𝑘
= (2 + 3𝑒 𝑗 4 ) ∑ 𝑒 2
𝑛

𝑛=0
𝜋𝑘 1 𝑗2𝜋𝑘
−𝑒
= (2 + 3𝑒 𝑗 4 ) 𝜋𝑘 if 𝑘 = 0,4 and 0 otherwise
𝑗
1−𝑒 2
𝜋𝑘
𝑗
= (2 + 3𝑒 4 ) × 4 if 𝑘 = 0,4 and 0 otherwise
Therefore, 𝑋(0) = 5 × 4 = 20, 𝑋(4) = (2 − 3) × 4 = −4
𝑋(𝑘) = {20,0,0,0, −4,0,0,0}

Ans c
4. The DFT of 𝑥1 (𝑛)𝑥2 (𝑛) is
 1
𝑋 (𝑘) ⊛ 𝑋2 (𝑘)
𝑁 1
Ans d
𝜋𝑛
5. Given 𝑥(𝑛) = cos ( 2 ) = {1,0, −1,0} for 0 ≤ 𝑛 ≤ 3 and ℎ(𝑛) = {2, −1,3, −2}. Let ⨀
denote element-wise product and sum. The circular convolution of 𝑥(𝑛) and ℎ(𝑛) is
evaluated as follows
𝑌(0) = {2, −1,3, −2}⨀{1,0, −1,0} = −1
𝑌(1) = {2, −1,3, −2}⨀{0,1,0, −1} = 1
𝑌(2) = {2, −1,3, −2}⨀{−1,0,1,0} = 1
𝑌(3) = {2, −1,3, −2}⨀{0, −1,0,1} = −1
The output is 𝑦(𝑛) = {−1,1,1, −1}
Ans a
6. Given 𝑥(𝑛) has DFT 𝑋(𝑘), then 𝑋(𝑛) has DFT
𝑁𝑥(−𝑘 mod 𝑁)
𝑁−1 𝑁−1
1 𝑘𝑛 1 𝑘𝑛
𝑥(𝑛) = ∑ 𝑋(𝑘)𝑒 𝑗2𝜋 𝑁 ⇒ 𝑥(𝑘) = ∑ 𝑋(𝑛)𝑒 𝑗2𝜋 𝑁
𝑁 𝑁
𝑘=0 𝑛=0
𝑁−1 𝑁−1
1 (𝑁−𝑘)𝑛 1 𝑘𝑛
⇒ 𝑥(𝑁 − 𝑘) = ∑ 𝑋(𝑛)𝑒 𝑗2𝜋 𝑁 = ∑ 𝑋(𝑛)𝑒 −𝑗2𝜋 𝑁
𝑁 𝑁
𝑛=0 𝑛=0
⇒ 𝑁𝑥(−𝑘 mod 𝑁) ↔ 𝑋(𝑛)
Ans b
𝜋𝑛
7. Given 𝑥(𝑛) = sin ( 2 ) = {0,1,0, −1} for 0 ≤ 𝑛 ≤ 3. ℎ(𝑛) = {4,2,1,3}. Let ⨀ denote
element wise product and sum. The circular convolution of 𝑥(𝑛) and ℎ(𝑛) is evaluated
as follows
𝑦(0) = {4,2,1,3}⨀{0, −1,0,1} = 1
𝑦(0) = {4,2,1,3}⨀{1,0, −1,0} = 3
𝑦(0) = {4,2,1,3}⨀{0,1,0, −1} = −1
𝑦(0) = {4,2,1,3}⨀{−1,0,1,0} = −3
Hence, 𝑦(𝑛) = {1,3, −1, −3}
Ans b
2𝜋
8. The 2 × 2 IDFT matrix is given as follows. 𝑊2 = 𝑒 −𝑗 2 = 𝑒 −𝑗𝜋 = −1. The DFT
matrix is
1 1 1 1 1 1
[ 1] = [ ]
𝑁 1 𝑊2 2 1 −1
Ans a
9. The 8 −pt of {4,0,0,0,6,0,0,0} is
4𝑘
𝑋(𝑘) = 4 + 6𝑒 −𝑗2𝜋 8 = 4 + 6𝑒 −𝑗𝜋𝑘 = {10, −2,10, −2,10, −2,10, −2}
Ans d
𝜋𝑛
10. Given 𝑥(𝑛) = cos ( 2 ) = {1,0, −1,0} for 0 ≤ 𝑛 ≤ 3 and ℎ(𝑛) = {−1,2, −3,4}. Let ⨀
denote element-wise product and sum. The circular convolution of 𝑥(𝑛) and ℎ(𝑛) is
evaluated as follows
𝑌(0) = {−1,2, −3,4}⨀{1,0, −1,0} = 2
𝑌(1) = {−1,2, −3,4}⨀{0,1,0, −1} = −2
𝑌(2) = {−1,2, −3,4}⨀{−1,0,1,0} = −2
 𝑌(3) = {−1,2, −3,4}⨀{0, −1,0,1} = 2
The output is 𝑦(𝑛) = { 2, −2, −2,2}
Ans c