Real/Complex Zeros for Filter Transfer Function

So I've been reading into filter design and came across how to determine the transfer function of a filter, when given it's poles and zeroes. I've come across the following question that has me a bit confused about how to determine the zeroes of a transfer function.

Suppose you have a third-order low-pass filter that has transmission zeroes at $$\\omega=2,\infty\$$, poles at $$\-1,-0.9\pm1.1j\$$ and a DC gain of unity. Going through the work, I get the following transfer function:

$$T(s)=\frac{K(s^{2}+4)}{(s+1)(s+0.9+1.1j)(s+0.9-1.1j)}$$

where $$\K\$$ is the gain factor (calculated using the fact that the DC gain is unity).

Here is my issue. The question states that there are multiple transmission zeros, but does not state how many zeros are at $$\\omega=2\$$. I assumed that the zero at $$\\omega=2\$$ was complex, leading to a conjugate pair $$\(s+2j)(s-2j)\$$ and hence how I got $$\s^{2}+4\$$ for the numerator. But is there someway to show that said zero is in fact a conjugate pair, and not real?

Suppose the question instead stated that there was "a transmission zero at $$\\omega=2\$$ ". Would the numerator of the transfer function then be equal to $$\s-2\$$, with a single real zero, since the zero must be real if there is only one of them (i.e. no conjugate pair)?

• How does that TF produce a zero at infinity? – Andy aka Mar 30 at 21:17
• electronics.stackexchange.com/questions/523446/… – Tony Stewart EE75 Mar 30 at 21:54
• Can the numerator be of the form $(s+2)(s+2) = s^2+4s+4$ ? – AJN Mar 31 at 12:40
• Can you post the question verbatim ? And a link to the question if access to the material you are reading is available online ? – AJN Mar 31 at 12:41
• If you multiplied out the denominator there would be real terms that are unconnected to s hence, there is no zero at infinity. OK, I guess they could cancel out with $s^2$ terms. I didn't do the math. – Andy aka Mar 31 at 15:05

Would the numerator of the transfer function then be equal to s−2

No, because S-2 implies Zero at $$(\sigma,\omega)=(2,0)$$ while you assumed Zero at $$\omega =2$$ and you can see in both cases imaginary parts are different.

The question states that there are multiple transmission zeros, but does not state how many zeros are at ω=2

only two possibilities are there either 1 Zero or 2 Zero at ω=2 , but two Zero at ω=2 is not possible because it leads to complex cofficient of numerator and which is not realizable and similarly two Zero at infinity is also not possible due to same reason as above.

I assumed that the zero at ω=2 was complex, leading to a conjugate pair (s+2j)(s−2j)

Your assumption seems correct let's see why-

Although Zero is given at ω=2 so one reasonable conclusion would be a zero at $$(\sigma,\omega)=(\sigma,2)$$ but it is also given that one Zero is infinity from which you can conclude that numerator is of 2nd degree polynomial . But for physically realizable filters one of the condition is that -all the coefficient of numerator and denominator should be real . Hence if one Zero is at $$S-(\sigma+j2)$$ so other would be at $$S-(\sigma-j2)$$. if we consider real part of Zero as a variable ,we cannot calculate its value because we have only one condition(DC gain) and two variables , so the only logical interpretation of $$\omega =2$$ is that its Zero is at (0,2) hence numerator will be $$(S-j2)(S+j2)=S^2+4$$

• Thank you for this explanation @user215805 . It seems my confusion was assuming that $\omega$ was the real part of the complex number, as opposed to the value of the imaginary part. – JTaft121 Mar 31 at 15:23
• The reason I asked about how things change if a single transmission zero was given actually comes from a version of that question where a second-order lowpass filter is considered. In this question, there are poles at $-0.9\pm1.1j$ and a transmission zero at $\omega=2$, with DC gain still unity. It also asks what the gain would be as $\omega$ approaches $\infty$. Would the numerator still be $s^{2}+4$? My initial guess was no because for a LP filter, gain falls to 0 as $\omega$ approaches $\infty$, which could only happen if deg(numerator)<deg(denominator). – JTaft121 Mar 31 at 15:25
• @JTaft121 , even if numerator is $$s^2+4$$ isn't still degree (numerator )<degree (denominator) and gain would be zero as as ω approaches ∞, And that is what you expected ? – user215805 Mar 31 at 16:32
• I expected the transfer function for the second order lowpass filter problem to be $$T(s)=K\frac{s^{2}+4}{(s-0.9+1.1j)(s-0.9-1.1j)}$$ since there is one transmission zero at $\omega=2$ and poles at $0.9\pm1.1j$. But now I am not sure because the transfer function I have given has deg(num)=deg(den), which does not go to 0 as $\omega$ approaches $\infty$. – JTaft121 Apr 1 at 15:07