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Hello All,
Wanted to ask you for some advice as I am a little stuck on a 1st order FIR low pass filter question
If the following input signal (which is made up of 2 components) is applied to the low pass filter:
\(x(t) = 5 + 12cos(20 \pi t) \)
\( H(e^{j\omega})=\frac{1+e^{-j\omega}}{2}\)
Using superposition the out signal can be determined. If we assume that \( Fs=40 Hz \) ,
\(x(t) = 5 + 12cos [\frac {2 \pi 10 }{40}] \)
\(x[n] = 5 + 12cos [\frac {\p}{2} n] \)
Output in discrete time is equal to the magnitude and phase response at both frequencies i.e.
\( \omega 1 = 0\) and \( \omega 2 = \frac{\pi}{2}\)
\( H(e^{j\omega 1})=\frac{1+e^{-j0}}{2}\)
\( |H(e^{j\omega 1})|= 1\)
\(\angle e^{j\omega1} = 0 degrees\)
\( H(e^{j\omega 2})=\frac{1+e^{-j\frac{\pi}{2}}}{2}\)
\( |H(e^{j\omega 2})|= \frac{1}{\sqrt{2}}\)
\(\angle e^{j\omega2} = -45 degrees\)
\(y[n] = | H(e^{j\omega 1})|*5 + | H(e^{j\omega 2})|*12cos [\frac {\p}{2} n - \angle e^{j\omega 2} \)
\(y[n] = 5 + \frac{1}{\sqrt{2}}*12cos [\frac {\p}{2} n - \frac{\pi}{4} \)
My question is how can I prove this, in MATLAB and/or on paper. These are the 2 ways I think
1) Determine the impulse response from the transfer function, I thinks its \( h[n] = \frac {1}{2}, \frac{1}{2} \), then using circular convolution to determine y[n].
2) Take the Z transform of the input signal, then multiply the 2 and take the inverse z transform.
I have used sptool in MATLAB, however failed to proove it that as well.
Sadly I can not do it using any of the 3 methods, hopefully I am not to far away from the solution.
Wanted to ask you for some advice as I am a little stuck on a 1st order FIR low pass filter question
If the following input signal (which is made up of 2 components) is applied to the low pass filter:
\(x(t) = 5 + 12cos(20 \pi t) \)
\( H(e^{j\omega})=\frac{1+e^{-j\omega}}{2}\)
Using superposition the out signal can be determined. If we assume that \( Fs=40 Hz \) ,
\(x(t) = 5 + 12cos [\frac {2 \pi 10 }{40}] \)
\(x[n] = 5 + 12cos [\frac {\p}{2} n] \)
Output in discrete time is equal to the magnitude and phase response at both frequencies i.e.
\( \omega 1 = 0\) and \( \omega 2 = \frac{\pi}{2}\)
\( H(e^{j\omega 1})=\frac{1+e^{-j0}}{2}\)
\( |H(e^{j\omega 1})|= 1\)
\(\angle e^{j\omega1} = 0 degrees\)
\( H(e^{j\omega 2})=\frac{1+e^{-j\frac{\pi}{2}}}{2}\)
\( |H(e^{j\omega 2})|= \frac{1}{\sqrt{2}}\)
\(\angle e^{j\omega2} = -45 degrees\)
\(y[n] = | H(e^{j\omega 1})|*5 + | H(e^{j\omega 2})|*12cos [\frac {\p}{2} n - \angle e^{j\omega 2} \)
\(y[n] = 5 + \frac{1}{\sqrt{2}}*12cos [\frac {\p}{2} n - \frac{\pi}{4} \)
My question is how can I prove this, in MATLAB and/or on paper. These are the 2 ways I think
1) Determine the impulse response from the transfer function, I thinks its \( h[n] = \frac {1}{2}, \frac{1}{2} \), then using circular convolution to determine y[n].
2) Take the Z transform of the input signal, then multiply the 2 and take the inverse z transform.
I have used sptool in MATLAB, however failed to proove it that as well.
Sadly I can not do it using any of the 3 methods, hopefully I am not to far away from the solution.
