How to find the magnitude and phase of a transfer function

The function is 1/ (R+jwRC) and the book I'm using said that the magnitude is 1/sqrt(1+(w/a)^2) were a=1/RC and the phase is -tan^-1(w/a). I have no idea how they got either answer and the book is no help. Please explain how to get it.

hello Philip,

you should ask question properly. for example explain what the circuit is. sure it is some RC circuit but there can be several different connections - even for just two components.

suppose we have low pass filter for example

then Z1=R, Z2=1/(jwC) and in case of high pass filter Z1=1/(jwC) while Z2=R

output voltage is:
Vo=Vi*Z2/(Z1+Z2 (direct result of voltage divider equation).

by definition H=Vo/Vi
so in this case
H=Z2/(Z1+Z2)

then for low pass filter we get

H=[1/(jwC)] /[R+1/(jwC)]

H=[(1/jwC)] / [(jwRC+1)/jwC]

and finally after cancelling denominators we get

H=1/[1+jwRC)

if we define a=1/(RC)

we get

H=1/(1+jw/a) etc.

we can calculate magnitude different ways but result is the same.
simplest to do by hand is to do

h=magnitude of denominator
then
|H|= 1/h

where

h=sqrt(1^2+(w/a)^2)

so

|H|=1/sqrt(1^2+(w/a)^2)

and
tan(alpha)=imaginary part / real part
but since this is in denominator, angle is negative (you need to keep track of this when doing shortcuts like this; otherwise, do the expanded form where H=a+jb ann in this case you will get positive angle
tan(angle=b/a).

-tan(alpha)=(w/a) / (1)
which means
alpha=-arctan(w/a)


to get expanded form, you can multiply H by one (there are only two tricks in math: multiply by one, and add zero).

in our case we have transfer function in form

H=1/(u+jv)

to get normal form we multiply it by one where

1=(u-jv)/(u-jv)

note that u-jv is conjugate of denominator.
since we are multiplying by 1, H does not change but the representation will become a+jb.