348: EE Lab 6, Spring 2008 Frequency Response of FIR Filters
1
Introduction
Lab 6 examines the frequency response of discrete-time FIR lters, the Discrete-Time Fourier Transform (DTFT, used for discrete non-periodic signals) and the frequency response of the output of an FIR lter. Your lab report is due at the beginning of class Tuesday of two weeks hence, April 15. Labs are to be done individually, although you may consult together in groups in regards to the lab. It is expected that you will need to spend some time outside of the scheduled lab period to complete and document each lab.
1.1
Lab Goals
The primary goals of this lab are: Write a program to calculate the DTFT of a discrete-time signal. Use the Matlab freqz function to calculate the frequency response of a discrete-time signal. Examine the frequency response H() of an FIR lter and the frequency response Y () of its output y[n] for a given input x[n].
2
2.1
Frequency Response
Discrete-Time Fourier Transform (DTFT)
The DTFT is used to nd the continuous frequency response X() of a non-periodic discrete-time signal x[n]. The frequency response X() is continuous and periodic in frequency, with period 2. The frequency response X() is found as
X() =
n=
x[n] exp(jn).
(1)
This equation is known as the DTFT synthesis equation, or simply the DTFT of x[n]. Note that for x[n] having nite range of support from N1 to N2 , the DTFT reduces to a nite number of terms, given by
N2
X() =
n=N1
x[n] exp(jn).
(2)
The inverse DTFT or IDTFT, or DTFT synthesis equation, is given by x[n] = 1 2
X() exp(jn)d.
=
(3)
Matlab provides a function to calculate the DTFT of a non-periodic discrete signal: the freqz function. The frequency response H(k) of an FIR lter with impulse response h[n] may be found over a frequency range of freqrange (often freqrange= : /100 : , or 0 : /100 : ) with the command freqz(h,1,freqrange).
1
EE 348: Lab 6, Spring 2008
2
2.2
Lab Exercise: DTFT
1. Write a Matlab function to calculate the DTFT of a non-periodic discrete signal x[n], assuming x[n] has nite range of support from N1 to N2 . Include your Matlab function in your lab report. You may write this DTFT function in C, if you prefer.
1 1 2. Consider the causal FIR lter from Lab 4, Section 2.1, with impulse response h[n] = 10 [n] + 10 [n 1] + 9 1 1 1 1 1 1 1 1 1 k=0 [nk]. 10 [n2]+ 10 [n3]+ 10 [n4]+ 10 [n5]+ 10 [n6]+ 10 [n7]+ 10 [n8]+ 10 [n9] = 10 This impulse response is discrete-time and nonperiodic. Use your Matlab function to calculate the DTFT H() for , with frequency steps of = /100. Plot both the magnitude |H()| and the phase H() vs , for . Label both plots and include in your lab report.
3. Now use the freqz function in Matlab to conrm your frequency response results for H(). Again, plot both the magnitude |H()| and the phase H() vs , for . Label both plots, indicating they result from the Matlab freqz function, and include in your lab report. 4. Based on the magnitude of lters frequency response |H()|, is the lter h[n] a lowpass, bandpass or highpass lter?
3
Frequency Response of Filter Output
For an input x[n] applied to an LTI lter with impulse response h[n], the output will be y[n] = h[n] x[n]. Remember that when we considered two LTI in systems cascade with impulse responses h1 [n] and h2 [n], we found that the overall impulse response h[n] = h1 [n]h2 [n], and the overall frequency response was H() = H1 ()H2 (). In other words, convolution in the time domain equals multiplication in the frequency domain. We can apply this principle to the lter output y[n], which is the convolution of h[n] and x[n]. By the convolution property of the Fourier Transforms, we know that convolution in time is equivalent to multiplication in frequency. In the frequency domain, the output frequency response Y () is found as Y () = H() X(). (4)
So we need to nd the frequency response of both the lter, H(), and the input, X(). Once the output frequency response Y () has been calculated, then the output time-domain response can be found from the IDTFT of equation 3. This sounds like a complicated process; however, multiplication in the frequency domain is often much easier than convolution in the time domain. The lab exercise illustrates this process.
3.1
Lab Exercise: Frequency Response of Filter Output
1. Use the same lter impulse response h[n] with frequency response H() found earlier in Section 2.2. 2. Generate a discrete non-periodic input of length 100 as x[n] = 1 + cos(n/10) + e[n] for n = 0, . . . 99, where e is a random noise vector, also of length 100. Generate e rst with the command e =0.5*randn(1,100), 2 which generates a Gaussian zero-mean random sequence of length 100 with variance e = 0.25. This input x[n] is non-periodic due to the addition of the random noise vector. Plot x[n] vs n, label and include in your lab report. 3. Find the input frequency response X() by using either your DTFT program or Matlabs freqz() function. Be sure to use the same range of for X() as you did for H(). Plot both the magnitude |X()| and phase X() for . Label both plots. Include both plots in your lab report.
EE 348: Lab 6, Spring 2008
3
4. Find the output frequency response...