# Check work: Expression for 3dB bandwidth for frequency response of high - pass filter having just one pole

The input output relationship is given:

$$y[n]=ay[n-1]+bx[n]$$

$$-1

I applied z-transform to get the transfer function :

$$H(z)=\frac{b}{1-az^{-1}}$$

Since $$-1, the maximum power will occur when $$w=\pi$$ where $$w$$ is frequency.

So max $$|H(z)|^2=\frac{b^2}{(1+a)^2}$$

At the 3dB frequency, the power will be half of this. So if the 3dB frequency is $$w$$, then:

$$\frac{b^2}{1+a^2-2acosw}=\frac{b^2}{2(1+a)^2}$$

$$1+a^2-2acosw=2(1+a)^2$$

Solving this I got

$$w=cos^{-1}(\frac{-1-a^2-4a}{2a})$$

Is this correct? If not, which step is wrong?

• It's correct, and also pretty straightforward to check. You can always plot the result for a few values of $a$ to see if it makes sense. Sep 25 '19 at 13:45
• @Matt L. Thanks. Do you mean plotting using a software or something? I haven't used any plotting softwares so far. Sep 25 '19 at 14:13
• If you want to do a bit of signal processing it's very handy to have some simulation software, such as SciPy or Octave/Matlab. Sep 25 '19 at 14:32