# Why is there a negative in front of the phase response equation for this complex exponential?

first time on here!

I'm working through "Digital Signal Processing using MATLAB" by Vinay and Proakis. Good book.

I am stuck on this example tho.

Shouldn't the imaginary part in the denominator (magnitude calculation) be (-0.9sin(w))^2 since the Frequency Response equation has an e^-jw?

Also why does the angle calculation include a -arctan instead of just a regular arctan.

Thanks

For the magnitude, the denominator term will be $$1-0.9e^{-j\omega}=1-0.9(cos\omega - jsin\omega) = 1-0.9cos\omega+0.9jsin\omega$$

Hence the plus sign.

For the phase, if you multiply the numerator and denominator with the complex conjugate,

$$H(e^{j\omega}) = \frac {1-0.9cos\omega-0.9jsin\omega}{(1-0.9cos\omega+0.9jsin\omega)(1-0.9cos\omega-0.9jsin\omega)}$$

The denominator will become real. Therefore, $$\angle H(e^{j\omega}) = arctan(\frac {-0.9 sin\omega}{1-0.9cos\omega})$$ Taking the minus sign outside is how you get the negative in the phase.

• Thanks very much! Makes sense now! – Dom May 10 '20 at 8:48
• Why would you want to multiply the numerator and denominator by the complex conjugate? Couldnt you just take the arctan of the imaginary part divided by the real part? – Dom May 10 '20 at 8:58
• When the complex number is represented as a+ib, you take the arctan of imag/real i.e b/a. The given complex number is in 1/a+ib form. To get the complex number in the a+ib form, the numerator and denominator were multiplied by the complex conjugate. – DSP Novice May 10 '20 at 9:32