Question

4. A signal x()3cos(4007t) is sampled at rate of 1200 Hz to compute its 36-point DFT X[k]. a) What is Xk]? b) Plot Xk]

Answer 1

The given signal is
r(t 3 cos(400Tt
and the sampling rate is
s=1200
Hz.

Let us determine the frequency of the given signal. The general expression of sinusoidal signal is

.

(tA cos(2Tft)

Equating with the given signal. We get,
VA=B
and
2Tf400
. Thus,
200
Hz.

Now, let us performing the sampling and obtain the sampled signal .

n

For sampling, is placed in the continuous signal.

t nT fs

Thus,

200 r[n]3 cos(2T- 1200 -n

r[n]3 cos(2-n 6

We need not compute 36 points of
n
for calculation of 36-point DFT . Because, it is time consuming and not suitable. Instead, we use the properties of DFT for computation.

Now, let us split the discrete cosine signal into individual exponential components.

We know that, $\cos \theta = \frac{1}{2} (e^{j\theta} + e^{-j\theta} )$

Thus, $x[n] = \frac{3}{2} e^{j 2\pi \frac{1}{6}n} + \frac{3}{2} e^{-j 2\pi \frac{1}{6}n}$

==============================================================

The N-point DFT is expressed as

$X[k] = \sum_{n=0}^{N-1} x[n]e^{-j2\pi \frac{k}{N}n}$

Placing N = 36 in the above expression, we get

35 Χ-Σ mle-/2r πr -j2 36 n-0

Replace with the individual components.

n

$X[k] = \sum_{n=0}^{35} (\frac{3}{2}e^{j2\pi \frac{1}{6}n} e^{-j2\pi \frac{k}{36}n} + \frac{3}{2}e^{-j2\pi \frac{1}{6}n}e^{-j2\pi \frac{k}{36}n})$

Grouping the powers in the exponential terms.

$X[k] = \sum_{n=0}^{35} (\frac{3}{2}e^{j2\pi (\frac{1}{6} - \frac{k}{36})n} + \frac{3}{2}e^{-j2\pi (\frac{1}{6} + \frac{k}{36})n})$

$X[k] = \frac{3}{2} \sum_{n=0}^{35} e^{j2\pi (\frac{1}{6} - \frac{k}{36})n} + \frac{3}{2} \sum_{n=0}^{35} e^{-j2\pi (\frac{1}{6} + \frac{k}{36})n}$

We know that DFT property,

$\sum_{n=0}^{N-1}e^{j2\pi (\frac{1}{6} - \frac{k}{36})n} = \left\{\begin{matrix} N & \frac{1}{6} = \frac{k}{36}\\ 0 & \text{otherwise} \end{matrix}\right.$

N-1 N 0 otherwise ej2(+n 36 n-0

The above summation expression yields and , respectively.

Because, $\frac{1}{6} = \frac{k}{36}$ , cross multiplying, $k = \frac{36}{6} = 6$

and  $\frac{1}{6} = -\frac{k}{36}$, cross multiplying,

36 k -6 6

Using the DFT properties, we know that DFT coefficients are non-zero at location and .

Thus, $X[6]= \frac{3}{2}*36 = 54$

Since, cannot be negative in the DFT coefficients, it represents,

$X[30]= \frac{3}{2}*36 = 54$

================================================================

The 36-point DFT coefficients are

X 0,0,0, 0, 0,0,54, 0,0,0, 0,0,0,0, 0,0,0,0, 0, 0,0,0,0,0,0,0, 0, 0,0,0,54,0,0, O, 0, 0}

================================================================

Let us plot x[n] for 36-point and compute X[k] and plot X[k] using MATLAB.

-------- START OF MATLAB CODE ----------

% Sampling frequency
fs = 1200;
% Signal frequency
f = 200;
% Number of points
N = 36;
% Generate index variables
n = 0:1:N-1;
% Generate the signal
x = 3*cos(2*pi*(f/fs)*n);
% Plot the discrete signal using stem function
figure,stem(n,x),title('Discrete time /sampled signal'),xlabel('Time index'),ylabel('Amplitude'),grid on
% Compute DFT using fft function
X = fft(x, N);
% Display the coefficients
disp(X);
% Plot the DFT coefficients using stem function
figure,stem(n,abs(X)),title('DFT coefficients'),xlabel('Coefficient index'),ylabel('Magnitude'),grid on

-------- END OF MATLAB CODE ---------------

Save the above code as 'sample.m' and run 'sample' command in MATLAB command window. Two figures are generated, the first figure is time domain signal and the second figure is DFT coefficients. Each command has a comment above the command for explanation.

Time domain signal is shown below,

DFT Coefficients plot is shown below,

Discrete time/sampled signal 3 G 2 1 -1 -2 -3 0 35 30 25 20 15 Time index 10 5 -O G -O -0 Amplitude

==================================================================

DFT coefficients 60 50 40 30 20 10 5 10 15 Coefficient index 20 25 30 35 Magnitude