Engineering Bandwidth Filtering: Get Help Solving Last Problem

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The discussion focuses on solving a DSP problem related to determining the 3dB bandwidth of two filters. The initial suggestion is to compute and plot the magnitude of each filter's transfer function. Participants emphasize the importance of showing work and clarifying the plotting method, particularly regarding the variable used. The final advice is to identify the upper and lower -3dB frequencies from the transfer functions provided. This approach has motivated the original poster to seek a solution effectively.
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Homework Statement
Determine the 3dB bandwidth of the filters

$$ H_1(z) = \frac{1-a}{1-az^{-1}} $$

$$ H_2(z) = \frac{1-a}{2} \frac{1+z^{-1}}{1-az^{-1}} $$

if 0<a<1
Relevant Equations
Not sure.
Hi everyone

I am finishing the last problem for a DSP problem set and just frankly have no idea where to start this one. I'm thinking I could compute the magnitude of each filter and then compare, but again am not sure. Can anyone point me in the right direction?

Thanks
 
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Try that first. Plot the magnitude like you were saying. You have to show some work or attempt before people can help you here.

edit:

When you plot it: Are you going to plot as a function of ##z## or as something else? Did the professor use anything else or give you an equation where ##z## was some function of something else that had ##\Omega## in it?
 
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snatchingthepi said:
Homework Statement:: Determine the 3dB bandwidth of the filters

$$ H_1(z) = \frac{1-a}{1-az^{-1}} $$

$$ H_2(z) = \frac{1-a}{2} \frac{1+z^{-1}}{1-az^{-1}} $$

if 0<a<1
Relevant Equations:: Not sure.

I am finishing the last problem for a DSP problem set and just frankly have no idea where to start this one. I'm thinking I could compute the magnitude of each filter and then compare, but again am not sure. Can anyone point me in the right direction?
You can just figure out where the upper and lower -3dB frequencies are from each transfer function...

1618423279074.png
 
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These posts were enough to encourage me to find a good answer. Thank you both.
 
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