h61554 s 00000/00051/00116 d D 1.10 24/10/24 12:52:16 bevans 10 9 c Updated e s 00001/00001/00166 d D 1.9 24/10/19 21:11:03 bevans 9 8 c Updated e s 00011/00003/00156 d D 1.8 21/10/22 21:14:45 bevans 8 7 c Updated e s 00022/00003/00137 d D 1.7 21/10/22 21:07:22 bevans 7 6 c Updated e s 00001/00000/00139 d D 1.6 21/10/22 20:43:27 bevans 6 5 c Updated e s 00008/00005/00131 d D 1.5 21/10/22 20:41:17 bevans 5 4 c Updated e s 00037/00003/00099 d D 1.4 21/10/21 16:47:53 bevans 4 3 c Updated e s 00030/00006/00072 d D 1.3 21/10/14 19:15:34 bevans 3 2 c Updated e s 00002/00000/00076 d D 1.2 21/10/14 18:24:22 bevans 2 1 c Updated e s 00076/00000/00000 d D 1.1 21/10/14 18:14:31 bevans 1 0 c date and time created 21/10/14 18:14:31 by bevans e u U f i f e 0 t T I 1 EE313 Linear Systems & Signals - Homework 6 Hints

EE313 Linear Systems & Signals - Homework 6 Hints

Homework #6 assignment

Problem 6.1. In homework problem 6.1(a), a linear time-invariant (LTI) system with E 8 I 8
Problem 6.1(a). A linear time-invariant (LTI) system with E 8 input signal x[n] and output signal y[n] is described by
y[n] = x[n] + x[n - 1] for n ≥ 0
As a necessary condition for LTI properties to hold, the LTI system must be at rest. This means that the initial condition x[-1] = 0.
The LTI system is a version of the two-point averaging filter in which the coefficients of 1/2 and 1/2 have been scaled by 2.
1. The impulse response of a system is the system response (output) when the input is an impulse signal. Denoting the impulse response as h[n] and letting the input signal be D 4 the impulse signal d[n] E 4 I 4 the impulse signal d[n], E 4
  h[n] = d[n] + d[n - 1] for n ≥ 0
  D 4
2. The step response is the response an input signal that is the unit step signal u[n]. The unit step signal is 1 when n ≥ 0 and 0 otherwise. E 4 I 4
3. The step response is the response an input signal that is the unit step signal u[n] where the unit step signal is 1 when n ≥ 0 and 0 otherwise. Let x[n] = u[n],
  y_step[n] = u[n] + u[n - 1] for n ≥ 0
4. The frequency response of the LTI system is the Fourier transform of the impulse response, where oo means infinity:
```
   j w       oo          -j w n    1         -j w n                -j w        -j w  
H(e   ) =   Sum   h[n] e       = Sum  h[n] e       = h[0] + h[1] e     = 1 + e     
          n = -oo                n=0                                               
```
  D 5 We can factor the frequency response into polar form per lecture slide 9-9: E 5 I 5 D 8 We can factor the frequency response into polar form per lecture slide 9-9 by E 8 I 8 We can factor the frequency response into magnitude-phase form per lecture slide 9-9 by E 8 factoring out the phase shift corresponding to the delay equal to the index of the midpoint of the impulse response, which is at an index of 1/2: E 5
```
   j w     -j (w/2)      j (w/2)    -j (w/2)      -j (w/2)               
H(e   ) = e          (  e        + e         ) = e          ( 2 cos(w/2) )

   j w                -j (w/2)
H(e   ) = 2 cos(w/2) e
```
  - Magnitude response: 2 cos(w/2) which is non-negative for -pi ≤ w < pi.
  - Phase response: -w/2 which is a line of slope -1/2 that intersects the origin.
5. The MATLAB command
```
D 5
freqz( [ 1 1] );
E 5
I 5
freqz( [1 1] );
E 5
```
  will plot the magnitude and phase response for an LTI system with impulse response d[n] + d[n - 1]. The magnitude response, by default, will be plotted in decibels vs. discrete-time frequency, where A_dB = 20 log₁₀ | A |. E 4
I 4 D 7 E 7 E 4
I 8
Problem 6.1(c). Please see the hints above for problem 6.1(a).
As in problem 6.1(a), we can factor the frequency response into magnitude-phase form per lecture slide 9-9 by factoring out the phase shift corresponding to the delay equal to the index of the midpoint of the impulse response, which is at an index of 1 and hence the phase shift is exp(-j w).
E 8 I 7
Problem 6.2(a). As we had seen on lecture slide 8-8 and in Signal Processing First Section 5-6, the output of an LTI system y[n] is the convolution of the input signal x[n] and the LTI system's impulse response h[n]
When we cascade two LTI systems with impulse responses h₁[n] and h₂[n], as we had seen on lecture slide 8-11 and in Signal Processing First Section 5-7,
y[n] = h₁[n] * h₂[n] * x[n]
We can write the input-output relationship for the cascade of two LTI systems as a single LTI system
y[n] = h[n] * x[n]
where
h[n] = h₁[n] * h₂[n] D 10
E 7 E 3
Problem 6.3. I 2 This problem is the same as problem 6.3 in fall 2018.
E 2 In homework problem 6.3(a), a linear time-invariant (LTI) system with D 3 input signalx[n] and output signal y[n] is described by E 3 I 3 input signal x[n] and output signal y[n] is described by E 3
D 3 y[n] = x[n] + x[n - 1] for n ≥ 0 E 3 I 3 y[n] = x[n] + x[n - 1] for n ≥ 0 E 3
As a necessary condition for LTI properties to hold, the LTI system must be at rest. This means that the initial condition x[-1] = 0.
The LTI system is a version of the two-point averaging filter in which the coefficients of 1/2 and 1/2 have been scaled by 2.
1. The impulse response of a system is the system response (output) when the input is an impulse signal. Denoting the impulse response as h[n] and letting the input signal be the impulse signal d[n]
  D 3 h[n] = d[n] + d[n - 1] for n ≥ E 3 I 3 h[n] = d[n] + d[n - 1] for n ≥ 0 E 3
  D 3
2. The transfer function in the z-domain is the z-transform of the impulse response h[n]: E 3 I 3
3. The transfer function in the z-domain is the z-transform of the impulse response h[n] : E 3
```
D 3
         oo          -n    1         -1                -1
E 3
I 3
         oo          -n    1         -n                -1
E 3
H(z) =   Sum   h[n] z   = Sum  h[n] z   = h[0] + h[1] z  
       n = -oo            n=0                            
```
4. To find the roots of the numerator (zeros) and the roots of the denominator (poles) of H(z),
```
            -1
H(z) = 1 + z  
```
  D 7 Zero is at z + 1 = 0 or z = -1. E 7 I 7 Zero is at z + 1 = 0 or z = -1. E 7
  D 7 Pole is at z = 0. E 7 I 7 Pole is at z = 0. E 7
5. The MATLAB code to plot the poles and zeros in the complex z plane is D 5
  D 3 h = [1 1]; E 3 I 3 h = [1 1];
  E 5 I 5
```
h = [1 1];
E 5
E 3
zplane(h);
D 5

E 5
I 5
```
  Note that zplane has the same syntax as freqz. E 5
I 6
E 6 Homework problems 6.3(b) and 6.3(c) are very similar. E 10

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bevans@ece.utexas.edu E 1