digital low pass Butterworth filter

A digital low pass Butterworth filter that has been designed using Bi-linear transformation has been a pole at $z=0.6$. It is also known that the filter's attenuate (at digital frequency) $\omega = 1.2$ is about $44$ dB. Find the filter order. Give at least one other pole of the digital filter (in Z domain).  • Now we know the assignment, but what's missing is your question about it. This is no homework site, so you're expected to share with us your efforts and to explain where you're stuck. May 3 '15 at 19:05
• I cant use any formula for filter design. So I am expecting some one to give me some hints on how to start. The passband and stop band ripple are not given May 3 '15 at 19:13
• A Butterworth filter has no passband or stopband ripples, it is maximally smooth, i.e. a Butterworth lowpass filter response decays monotonically. Do you know anything about Butterworth filters? If not, I guess you have to read up the basics before tackling such a problem. We give answers to specific questions, but it's useless to repeat all properties of Butterworth filters here. However, I'll post a few hints later on. If they don't help it means that you don't yet have enough background knowledge. May 3 '15 at 19:20

I'll try to give you some hints to get you started. First of all, you should know that the bilinear transform is given by

$$s=k\frac{z-1}{z+1}\tag{1}$$

If the analog prototype filter is normalized such that its cut-off frequency is $\Omega_c=1$, then the constant $k$ in (1) is given by

$$k=\frac{1}{\tan\left(\frac{\omega_0}{2}\right)}\tag{2}$$

where $\omega_0$ is the desired cut-off frequency of the discrete-time filter. Some more background on this is given in this answer.

You should also know that the poles of the normalized analog filter lie on a circle with radius $1$ centered at $s=0$ (of course the poles only lie on the left half of the circle). If the filter order is odd as in your example (why?), there must be a pole at $s=-1$. Now you can try to figure out how that pole is transformed to the $z$-plane, i.e.

$$\frac{1}{1+s}{\huge|}_{s=k\frac{z-1}{z+1}}=\ldots\tag{3}$$

From (3), knowing that that pole is transformed to a pole at $z=0.6$, you can determine the constant $k$, and from that, via Eq. (2), you can compute $\omega_0$. Now you know that the filter's attenuation is $3\text{ dB}$ at $\omega_0$, and it is $44\text{ dB}$ at $\omega=1.2$.

The last thing you should know is that with a Butterworth lowpass filter of order $N$, you get approximately $6N\text{ dB}$ attenuation per octave, at least in the frequency range considered here. Now figure out how many octaves there are between $\omega_0$ and $\omega=1.2$, figure out how many dB's difference in attenuation there must be between those two frequencies, and from this compute an estimate of the filter order $N$.

• Thank you very much. This helps me a lot to solve the problem. I found the order to be 3. In the middle of your explanation you mention that '..If the filter order is odd as in your example ..". Can you tell me how you come up with this statement. Thanks May 3 '15 at 23:30
• @GoitomHasa: The filter order must be odd because you have one real-valued pole at $z=0.6$. The other poles must come in complex conjugate pairs, so your total number of poles (= order) is $1$ (real-valued pole) + $2n$, where $n$ is the number of complex conjugate pole pairs. If the filter order were even, there would only be complex conjugate pole pairs and you wouldn't have a real-valued pole. May 4 '15 at 10:17
• @GoitomHasa: BTW, how did you come up with a filter order of $3$? It's not correct. May 4 '15 at 10:20
• Look my work which I uploaded it. I couldn't write all the equations in this website. Thanks May 4 '15 at 15:02