1

1. Linearity

We have two linear systems,

and

(a) What do these systems mean? How do you define these systems in Matlab?

(b) Show that for any input x[n], cascade connection of h[n] and g[n] in any order yields the same output. Does the same property hold for nonlinear systems? If not, give an example.

Answers:

(a) h[n] and g[n] can be considered as impulse responses of two linear systems, i.e. as outputs of those linear systems for an impulse in their input. Thus the outputs of these systems are linear combinations of present and previous input values. On the other hand, these systems can be considered as finite impulse response (FIR) filters. You can define these systems in Matlab in the following way:

h = [2 5 8 9];

g = [1 2 1];

(b) Proof: For any cascade connection of two or more linear time-invariant systems, the overall impulse response of the cascade can be obtained by convolving the impulse responses of all the subsystems. On the other hand, convolution is generally commutative operation, i.e. h[n]*g[n] = g[n]*h[n]. Therefore, because the impulse responses of the two cascades are the same, their outputs are the same for any input x[n].

This property (commmutativity of subsystems) does not generally hold for nonlinear systems although it holds in some special cases (e.g. if h[n] = abs(x[n]) and g[n] = kx[n]). For example, let h[n] = 2x[n] and g[n] = (x[n])², where x[n] is the input of the system. In case h[n] precedes g[n], total output will be y₁[n] = 4(x[n])². However, if g[n] precedes h[n], total output will be y₂[n] = 2(x[n])² ≠ y₁[n].

2. Complex plane

(a) Sketch the following numbers in the complex plane: i) 5, ii) , iii) . How can you represent these numbers in the form ?

(b) Assume that these numbers represent the response of a system to a sine wave. Explain in plain words how these signals change the input signal.

Answers:

(a) We can use Euler’s formula (e^iφ = cosφ + isinφ) to present these numbers in form a + ib

Figure 1: The given numbers in the complex plane

(b) The first system amplifies the signal with factor 5, the second (response to a sine wave is ) amplifies the signal with factor 2 and the phase of the signal changes with value 0.1p. The last system (response to a sine wave is ) multiplies signal by 3 and produces phase reversal, i.e. the phase of the signal changes with value p. (Actually, this means that the signal is multiplied by -3.)

By using the equations above, we get

3. Sampled signals

(a) In practical applications, sampled signals are typically represented as arrays of real numbers. In addition to the sample values themselves, what do you need to know before the properties of a sampled signal are uniquely defined?

(b) Assume a sampling frequency of f_s= 1000 Hz for this signal, and make it represent a 30 Hz sine wave. Plot both the analog and the digitized (sampled) sine wave with Matlab (Use the Matlab commands plot and stem). Now make a frequency vector which contains the frequencies from 0 up to 10000 Hz. Plot the frequency content of the sampled sine wave for this range of frequencies (Use the Matlab command freqz(sig,1,f,fs), where f is a frequency vector and fs is the sampling frequency). Based on the plot, what can you say about the frequency content of the sampled signal? Note that in Matlab, we always deal with sampled signals.

Answers:

(a) Properties of a sampled signal are uniquely defined if we know the sample values and sampling frequency.

(b) You can make a 30 Hz sine signal (sampling frequency F_s = 1000) in Matlab:

Fs = 1000; % sampling frequency

n = 0:999 % 1000 samples of a sine signal (1 second)

f = 30; % frequency of the sine signal

sine30_signal = sin(2*pi*f/Fs.*n);

The analog signal

plot(n/Fs,sine30_signal);

title(‘The 30 Hz sine signal (analog)’);

xlabel(‘Time (s)’);

The digitized (sampled) sine signal

stem(sine30_signal)

title(‘The 30 Hz sine signal (digitized)’);

xlabel(‘Samples (n)’);

The frequency content

f = 0:10000;

freqz(sine30_signal,1,f,Fs);

Figure 2: The analog (top) and ditized (bottom) 30 Hz sine signal.

Figure 3: The frequency content of the 30 Hz digitized sine signal,the sampling

rate is 1000 Hz.

From figure 3 we can see that the digitized signal contains also the frequencies 970 Hz, 1030Hz, 1970 Hz, 2030 Hz etc. (Discrete frequencies )

4. Convolution

Take the signal

and the system

(a) Filter with and call it.

(b) Find the frequency response of and . Then find the product of these frequency responses.

(c) Now compute the frequency response of. Compare the frequency response of with the product of the frequency responses of and.

Answers:

(a) Filtering (convolution)

x = [1 5 8 9];

h = [5 6 7];

y = conv(h,x);

(b) Frequency responses of x[n] and h[n]

[X,W] = freqz(x,1,512); % frequency response of x[n] calculated in 512 points

[H,W] = freqz(h,1,512); % frequency response of h[n] calculated in 512 points

XH = X.*H; % product of X and H

[Y,W] = freqz(y,1,512) % frequency response of y[n] calculated in 512

% points

plot(W/pi,abs(XH));

hold;

plot(W/pi,abs(Y),’r’);

Figure 4: X(eⁱ^w)H(eⁱ^w) (top) and Y(eⁱ^w) (bottom).

From figure 4 we can see that X(eⁱ^w)H(eⁱ^w) is equal to Y(eⁱ^w). (Convolution in time-domain is equal to product in frequency-domain)

5. Frequency representation of discrete signals

Take the signals

where . Choose a couple of frequencies and and find and for k=1,2. Did you notice any difference? Based on this, explain what range of the frequency response of a real signal is enough to give us information about its whole frequency response? (Note that you also have to deal with the frequencies like -0.7p and 5.3p).

Answers:

First, calculate frequency response of the signals and :

Z-Transforms:

Using the substitution , we get frequency responses:

If we choose frequencies , , , , ,and , we get:

and

In Matlab:

x1 = [1 2 3];

x2 = [1 2+3*i 3-4*i]

[X1,W] = freqz(x1,1,-4*pi:0.01:4*pi) % we calculate X between frequencies % -4*pi and 4*pi

[X2,W] = freqz(x2,1,-4*pi:0.01:4*pi) % we calculate X between frequencies % -4*pi and 4*pi

figure(1);

plot(W/pi,abs(X1));

title(‘The frequency response of the signal x1 (real)’);

ylabel(‘Amplitude’);

xlabel(‘Frequency ( 1 = pi)’);

figure(2);

plot(W/pi,abs(X2));

title(‘The frequency response of the signal x2 (complex)’);

ylabel(‘Amplitude’);

xlabel(‘Frequency ( 1 = pi)’);

Figure 5: The frequency response of the real signal x₁ (top) and the

frequency response of the complex signal x₂ (bottom).

From figure 5 we can see the difference between the amplitude responses of real and complex signals. The frequency response of the real signal is symmetric at zero frequency, The frequency response of the complex signal is not, although it is periodic, too. For the real signal, the frequency range enough is from 0 to p. For complex signal, we need the frequencies between 0 and 2p.