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Solution 3

1. The document discusses properties of Fourier transforms and signals in the frequency domain. It provides examples of Fourier transforms of common signals like sinc functions and derivatives of pulses. 2. Inequalities relating the Fourier transform magnitude of a signal to its derivative are derived. It is shown that the Fourier transform magnitude is bounded above by the integral of the signal's derivative. 3. For an amplitude modulated signal created by multiplying a carrier wave by a modulating signal plus noise, the document derives an expression for the amplitude sensitivity and shows that unwanted frequency components can be removed by filtering.

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57 views4 pages

Solution 3

1. The document discusses properties of Fourier transforms and signals in the frequency domain. It provides examples of Fourier transforms of common signals like sinc functions and derivatives of pulses. 2. Inequalities relating the Fourier transform magnitude of a signal to its derivative are derived. It is shown that the Fourier transform magnitude is bounded above by the integral of the signal's derivative. 3. For an amplitude modulated signal created by multiplying a carrier wave by a modulating signal plus noise, the document derives an expression for the amplitude sensitivity and shows that unwanted frequency components can be removed by filtering.

Uploaded by

Rehan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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1.

We have
 
 
F e j 2fot   h e j 2fo t  d  e j 2fot  h e  j 2fo d  e j 2fot H  f o 
 

Hence, xt   e j 2fot is an eigenfunction of the system

Ans a

2. pT(t) has a Fourier transform of Tsinc(fT). Hence, by duality Tsinc(tT) has Fourier
transform pT(-f) = pT(f). Given signal is sinc(2t). From above, 2sinc(2t) has Fourier
transform p2(f). Hence, sinc(2t) has Fourier transform ½ p2(f) i.e. ½ for |f|  1 and 0
otherwise
Ans c
3. The first derivative is pulse of height 1 from t = -2 to t = -1 and height -1 from t =1 to t
 3  3
=2 i.e.  p1  t    p1  t   , where pT t  = 1 for |t|  T/2 and 0 otherwise.
 2  2
Ans a

4. The second derivative is impulses with scaling 1 at t = -2, 2 and impulses with scaling -1
at t = -1,1. Hence, second derivative is
 t  2   t  2   t  1   t  1
Ans b

d 2 g t 
  j 2f  G f  . As can be seen above,
2
5. For this we use the property 2
dt
d 2 g t 
  t  2   t  2   t  1   t  1 . Its Fourier transform is,
dt 2
e  j 4f  e j 4f  e  j 2f  e j 2f  2 cos4f   2 cos2f 
  j 2f  G  f   2 cos 4f   2 cos2f 
2

cos2f   cos4f 
 G f  
2 2 f 2
Ans c
6. Inequality can be derived as shown below. We have,
dg t 
g t   G  f     j 2f G  f 
dt
dg t  j 2ft dg t  j 2ft dg t 
  
  j 2f G  f    dt e dt   dt e dt   dt dt
dg t 

dt dt
 G f   
2 f
dg t 

dt = 2. Hence, it follows that, G f  
1
Observe 

dt  f

Ans c

7. Consider the square-law characteristic v2 t   a1v1 t   a2v12 t  where a1, a2 are constants.
Let v1 t   Ac cos2f ct   mt  . Therefore using above equation and expanding terms,
v2 t   a1  Ac cos2f ct   mt   a2  Ac cos2f ct   mt 
2

 2a 
 a1 Ac 1  2 mt  cos2f ct   a1mt   a2 m 2 t   a2 Ac2 cos 2 2f ct 
 a1 
The first term at carrier frequency fc is the desired AM signal. The remaining terms are
2a
removed by filtering. Therefore, amplitude sensitivity is ka  2 .
a1
Ans c
8. Let the modulating wave m(t) be limited to the band -W  f  W as shown in figure
below. Then, from equation above we find that the amplitude spectrum |V2(f)| is as shown
in the next figure. It follows therefore that the unwanted terms may be removed from v2(t)
by designing the tuned filter at the modulator output to have a mid-band frequency fc and
bandwidth 2W which satisfy the requirement fc -W> 2W  fc > 3W. Note that this arises
since there is a baseband m2(t) term which has bandwidth 2W.

Ans b
9. The signal spectrum at points a, b is given in figure below. It can be readily seen from the
signal spectrum at point b that the required channel bandwidth for communicating the
composite message signal is 7.5 KHz.

Ans a
10. The signal spectrum at points a, b, c is given in figure below. It can be readily seen from
the signal spectrum at point c that the required channel bandwidth for communicating the
composite message signal is 17.5 KHz -2.5 KHz = 15 KHz.
Ans b

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