# E12 Assignment 11

Description:

Note: the notation on problem 2 was changed Wednesday April 8, 2015. The method of solution is unchanged.
Problem 5b was also changed.

Problems:
1. Convolution Property of Fourier Transform

Prove the convolution property of the Fourier Transform.  Hint: Start with definition of Fourier Transform of convolution and exchange order of integrals.

Solution:

2. Fourier Transform of Trapezoid as sum of triangles

a) Find the Fourier Transform, X(ω), for the function x(t) shown below.

This can be done without integration if you use the fact that a trapezoid is the sum of two triangular functions.  A triangular function of width 2·Tp and height 1, Λ(t/Tp), is shown below.

Its Fourier Transform is given by .

b) What are the Fourier Transform coefficients, cn, and the the Fourier Transform, XT(ω), of a signal, xT(t), that is a periodic extension of the one shown above.  Period=T, T>2·T1. If you write the Fourier Series expansion of xT(t), it is easy to find XT(ω). You don't need to simplify, but your result should be in terms of sinc() functions.

Solution:

a) We represent the function as the sum of two triangle waves of different widths, and slopes of magnitude 1/(T1-T0)  (i.e., rise/run),  so x(t)=f1(t)+f2(t).

b)

3. Fourier Transform of Two Triangles

Find the Fourier Transform, Y(ω), for the function, y(t), shown below.

Solution:

This function is simply the a trapezoidal pulse minus a rectangular pulse.   From a previous problem we know that the Fourier Transform of a trapezoidal pulse with value 1 for |t|<T0 and sloping to zero at |t|=T1 is given by:

and the Fourier Transform of a rectangular pulse (of height 1 and width 2T0)  is given by:

So

4. Fourier Xform of Half-wave Rectified Sine

This problem requires no integration.

a) For the function, p(t), what is P(ω)?

b) For the function pd(t), what is Pd(ω)?

c) For the function s(t)=sin(2π60t), what is S(ω)?  You can leave the result in terms of T and ω0=2π/T.

d) Find Y(ω) for the aperiodic function y(t) shown below.  You can express the result as the sum of two sinc functions.  You can leave the result in terms of T and ω0=2π/T.

Solution:

a)

b)

c)

d) Function is product of sine and shifted pulse (pulse has width T/2 and time shift of T/4), so the result is convolution of their Fourier Transforms. Recall that the convolution of a function with a shifted impulse yields a shifted version of that function.

5. Use Fourier XForm to find Fourier Series of Half Wave Rectified Sine

A function y(t) and its periodic extension yT(t) are shown below.

a) Find an expression for the Fourier Series coefficients of yT(t), cn.  You needn't try to simplify once you have it in terms of sinc() functions, but do make simplifications based on the identity ω0T=2·π.

Note: it can be shown (you needn't do this, but you may have done this in a previous problem set via a much more complicated procedure using direct integration) that the Fourier Series coefficients of yT(t) are also given by

b) Verify the value of c1, i.e., show that your result is equal to the one above.

Solution:

For n odd, and |n|≠1, cn=0 because arguments to sinc terms are integers.  This agrees with problem statement (e-jπn+1=0, for n odd).

otherwise

6. Fourier Transform of Rectangular Pulse via Laplace

Find the Fourier Transform of a rectangular pulse that is zero everywhere except between t=0 and t=T0 where it has a height of one:

x(t)=γ(t)-γ(t-T0)

a) by using a table of Fourier Transforms and Properties (this is just a rectangular pulse function of width T0 that is not centered on the origin).

b) by finding its Laplace Transform, X(s),  and then using the fact that the Fourier Transform of an absolutely integrable causal function is just the Laplace Transform  evaluated at s=jω.  (Note: absolutely integrable means ; causal means x(t)=0 for t<0.)

c) Show that the two are equal.   It may be useful to use a technique called "removing a half angle" (as well as Euler's identities for trig functions):

There is nothing complicated about this identity.  If you perform the multiplication on the right you can easily see that it equals the term on the left.

Solution:

a) This is just a pulse function of width T0 delayed by time T0/2.

b,c)  Find Laplace Transform and replace s by jω to get Fourier Transform, and then simplify.

7. Aliasing (straightforward)

Engineer A is taking some data from a strain gauge and is worried about electrical interference at 60 Hz so she takes data at 150 Hz to try to avoid problems with aliasing of the 60 Hz signal.  (Sampling at 150 Hz allows for representation of signals up to 75 Hz.)  Apparently Engineer A is unaware that there is also a lot of noise at 120 Hz (partly due to the fact that the 60 Hz is often rectified by power supplies - remember Fourier Series?).  To what frequency is the 120 Hz signal aliased? Note: the 60 Hz signal is not aliased, and need not be considered to solve this problem.

Solution:

Original freqs : -120 120 (thicker lines in image)
Image freqs : -330 -270 -180 -30 30 180 270 330
Aliased freqs : -30 30

8. Aliasing Demo

Thr graph below shows a 780 Hz cosine signal (solid blue line) . It also shows samples (circles) taken with a sampling frequency of 500 Hz. The 780 Hz signal aliases to 220 Hz (dashed line)

a) Explain why 780 Hz aliases to 220 Hz.

b) What other frequencies alias to 220 Hz?

c) Pick one of the frequencies from part b, and change the MATLAB code to verify your answer. There is no need to include the code, just the graph. The only line you should shange is "fsig = 780; % Signal frequency in Hz."

```t=linspace(0,0.01,1000);
fsig = 780;  % Signal frequency in Hz.

Fs = 500;  % Sampling frequency (Hz).
Ts = 1/Fs; % Sampling interval
tsamp = 0:Ts:0.01;  % Times of samples.

fAlias = 220;

plot(t,cos(2*pi*fsig*t)); hold on;
plot(tsamp,cos(2*pi*fsig*tsamp),'o');
plot(t,cos(2*pi*fAlias*t),'-.','Linewidth',2);  hold off;
legend({['Original signal: ' num2str(fsig) 'Hz'], ...
'Sampled Signal', ...
['Aliased signal: ' num2str(fAlias) 'Hz']}, ...
'Location','eastoutside');
xlabel('time');
ylabel('signal');
title('Signal: original, sampled and aliased');
```
Solution:

The cosine has frequency components at ±1720 Hz, and will have aliases at Fs·N ± fsig. (where Fs=sampling frequency, N=integer, and fsig=signal frequency).

a) 220 = 500·2 - fsig    (N=2)

b) fAlias = Fs·N ± fsig

c) See below for some answers (in order N=1, N=-1, N=-3)