The previous post in the series of articles on DSP was telling us about how you can get the frequency response of the LPF (low-pass filter) using the “method of the Laplace transform” (as I named it there). In that same article, it was said that there is another way to obtain formulae for the amplitude and phase response. In this article we will talk about a method called “direct deduction method” (a term is amateur and coined by me:-).

Let us recall the RC filter circuit:

1.SchemeThe ratio of the output voltage to the voltage at the input can be written as:

\frac{U_{out}}{U_{in}}=\frac{X_{c}}{X_{c}+R}

where resistance (impedance) of the capacitor in a particular frequency is:

X_{c}=\frac{1}{j{\omega}C}

We substitute this expression into the first formula:

\frac{U_{out}}{U_{in}}=\frac{X_{c}}{X_{c}+R}=\frac{1}{j{\omega}C\cdot({\frac{1}{j{\omega}C}+R})}=\frac{1}{1+j{\omega}RC}

What we see in this formula? You know, exactly this is the formula of the frequency response for the LPF! Are you doubting? Why? The formula really shows the ratio of the output voltage to the input voltage and this ratio depends on:

  • value of the resistor;
  • value of the capacitance of the capacitor;
  • the value of the complex frequency j\omega :-(.

Complex frequency ?! Again something strange … I think I have already said that sometimes having the final formula of the complex variable is much simpler and more convenient than if we use classical math without complex variables. Well, we shouldn’t be afraid of complex variable. It is not difficult thing! It’s just j=\sqrt{-1}. Frequently work with it boils down to getting the absolute value and phase. Absolute value for the resulting expression is equal to:

|K(\omega)|=|\frac{U_{out}}{U_{in}}|=\sqrt{{Re(\frac{1}{1+j{\omega}RC})}^2+{Im(\frac{1}{1+j{\omega}RC})}^2}

Who does not understand how to get the real and imaginary parts of a complex number (expression), I can explain it in such a way:

\frac{U_{out}}{U_{in}}=\frac{1}{1+j{\omega}RC}=\frac{1{\cdot}(1-j{\omega}RC)}{(1+j{\omega}RC)(1-j{\omega}RC)}=\frac{1}{1+({\omega}RC)^2}+j\frac{-{\omega}RC}{1+({\omega}RC)^2}=Re+jIm

Now, lets substitute the real and the imaginary part into the expression for |K(\omega)|:

|K(\omega)|=\sqrt{(Re(\frac{1}{1+j{\omega}RC}))^2+(Im(\frac{1}{1+j{\omega}RC}))^2}=\sqrt{\frac{1}{(1+({\omega}RC)^2)^2}+\frac{1}{1+({\omega}RC)^2}}

After simplification we get:

|K(\omega)|=\frac{1}{\sqrt{1+({\omega}RC)^2}}

The resulting expression is an amplitude response of the LPF filter. Drawing a plot:

Rendered by QuickLaTeX.com

Let us consider the phase response.

\varphi(\omega)=atan(\frac{Im(\frac{1}{1+j{\omega}RC})}{Re(\frac{1}{1+j{\omega}RC})})=atan(\frac{{-\omega}RC{\cdot}(1+({\omega}RC)^2)}{(1+({\omega}RC)^2){\cdot}1})

After reduction we obtain:

\varphi(\omega)=atan({-\omega}RC)

Phase response can be represented graphically as follows:

Rendered by QuickLaTeX.com

This graph shows that the higher the frequency of the signal at the input of the filter, the more it will be distorted (delayed) in phase (DC component is not distorted in phase distortion). Exactly because of this property of infinite impulse response IIR filters becomes not so attractive. But again, that’s another story …

As you can see, we got the same formulae and graphics, as in the previous article (although used two completely different methods of obtaining the final result).  Here is the power of physics and mathematics!