Electronic – Finding the amplitude response of an LTI system

signalsystem

I'm asked to find the amplitude response of an LTI system described by the difference equation:

$$
y[n] – ay[n-1] = x[n]
$$

After taking the Fourier transform of both sides and some math, I get the frequency response:

$$
H(e^{jw}) = 1/(1-ae^{-jw})
$$

But how do I get the amplitude response from this? I know the amplitude response is

$$ |H(e^{jw})| $$

but I'm unsure how to "get the length" of my frequency response. I have a feeling this is a simple math problem that I'm misunderstanding.

(Disclaimer: this is a question on a past final that I'm using to study. I have the answer to the question (and will post if proof is needed), but I want to understand the method)

Best Answer

Although you are in frequency domain you still should be able to get all parameters as you were in time domain. They are different domains but they both should represent the same thing. Time domain represent things in terms of amplitude in respect to time. Frequency domain represent things in terms of amplitude AND PHASE in respect to frequency values. Note that you should have both amplitude and phase in frequency domain, since in the time domain the phase can be represented in the same plot by a shift.

One way to represent these things in frequency domain is by dealing with complex numbers. Complex numbers can be viewed as vectors in a 2D space which have a length (as you said) and an angle. The length represents the output/input ratio and the angle represent the phase shift in comparison also to the input.

So, answering your question, you should calculate the H length to find your output/input ratio. To help you, imagine that:

\$e^{jw}=cos(w)+jsin(w)\$

In other words, its a complex number with always length of 1 and angle \$w\$

You can solve this by two methods:

-Vector method:

imagine that number 1 is \$Z=1+0.i\$ which is a vector to the right, with length 1 and angle \$0\$.

Imagine that \$e^{jw}\$ is a vector that I showed right above

Now add them. Then divive vectors 1 by the vector that you've found.

-Cartesian Coordinates:

represent all in terms of \$Z=a+jb\$ and also \$e^{jw}=cos(w)+jsin(w)\$

and imagine that you have:

\$\large Z = \frac{Z_1}{Z_2+Z_3}\$

and then find length of Z by:

\$|Z| = \sqrt{a^2+b^2}\$

Related Solutions

Amplitude and phase spectrum in MATLAB

This probably fits dsp.stackexchange.com better.

This is the most simple FT that you should understand. The basic rule is: a sine of frequency f0 in time domain is a delta at f0 and -f0 in frequency. You can do this theoretically, as shown in http://mathworld.wolfram.com/FourierTransformCosine.html.

Those peaks in your graph are consistent with this. However, it isn't true that there is a peak in at 90Hz. The peaks should be at 10 Hz and -10 Hz. There are two possible approaches here:

a) Use fftshift to correct this issue. This will take each frequency to its real place.

Fs=100; %sampling frequency
T=1; %signal length
N=T*Fs;     %number of samples
f=-Fs/2:Fs/N:Fs/2-Fs/N; %frequency vector
x=cos(2*pi*10*t);
Xf=fft(x);
Xf=fftshift(Xf);
plot(f,abs(Xf)) %magnitude

b) Assume the signal is real, so the spectrum is symmetric. It is true in this case, but it doesn't have to. You could just plot the first half of your fft, i.e. from 0 Hz to 50 Hz.

How to get the phase response of an arbitrary LTI system

I'm assuming you are trying to get a bode plot which shows the magnitude and phase response you would get if you were to test the the system by slowly sweeping the frequency a constant amplitude sine wave input and measuring the output amplitude and phase at various points.

If this is the case then there is only one frequency \$ \omega \$. Now we know that \$ e^{j \theta} = \cos(\theta) + i \cdot sin(\theta)\$.

You can therefore treat each exponential term as a vector with amplitude and phase and add them as vectors.

Note however, that if you compare the results with a real digital filter you will not get exactly the same result because the digital filter is only taking measurements at certain instants in time (the sampling frequency) and the data is quantised. However providing the sampling frequency is much larger than the frequency of interest it is a reasonable approximation.

Best Answer

Related Solutions

Amplitude and phase spectrum in MATLAB

How to get the phase response of an arbitrary LTI system

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