Help drawing Bode plots from given network function

Thread Starter


Joined Feb 12, 2015
Hello, the question is asking to sketch the magnitude Bode plot of the following network function:

\(H(\omega)=\frac{4(5 + j\omega)}{1 + j\frac{\omega}{50}}\)

My first problem is identifying what exactly the Bode plot will be plotting with regards to each axis; I can say that you will be plotting the log gain by log \(\omega\), but I can't really identify what those are with what I am given. However, I still tried to get somewhere, and what I came up with were these few steps:
factor out the five in the numerator to give us
\(H(\omega)=20\frac{(1 + j\frac{\omega}{5})}{1 + j\frac{\omega}{50}}\)
I did this because after looking at other example work, doing something along these lines would get it in a more general form, but I don't really have a good reason as to doing that other than other examples looked similar to that, and thats a fairly weak basis to go off.

I wish I could post a little more about what I know to help others help me, but I'm not sure what else to say at the moment. But, I do know that I don't know how to find the values that need plotting, so hopefully that is a fine starting point. Thanks in advance to anyone who can help!

P.S.- If I have multiple questions, but they aren't related, is it better to make multiple threads or just put them into one place?


Joined Mar 6, 2009
It's difficult to respond to your questions since there would be a number of things one would have to be aware of in relation to your overall understanding of the various underlying concepts. Such as your understanding of the meaning and manipulation of expressions involving complex numbers.
If you look at your function H(jw), it is an expression that describes a one-to-one relationship between the observed response at the particular network output terminals for a given sinusoidal stimulus applied to the network input terminals. If the input variable amplitude, frequency and phase are known then one can also state unambiguously what the output will be in those same terms.
The Bode plot shows what the input-output relationship (phase and magnitude) will be for a stated range of frequencies. The normalized form you show in the second expression of H(w) is useful in that it immediately informs you where the pole and zero "break" point frequencies lie along the independent frequency axis of the Bode plot.
I suggest you at least try to draw the Bode Plot for the expression and come back for further discussion. Post your answers please.
Last edited:


Joined Mar 31, 2012
You have an expression for the gain as a function of ω, so pick a value for ω and find the gain at that value and then you take the log (base 10) of each and plot the resulting point (on linear graph paper). Or, you can plot the log of the gain against ω directly on semi-log paper, or you could plot the gain against ω on log-log paper. Your choice.

The reason that the second form is the normalized form is that you can more easily see the frequencies at which the behavior of the graph changes. The term

\left( 1+j\frac{\omega}{\omega_0}\right)

is dominated by the value 1 for ω << ωo but is proportional to ω for ω>>ωo. So if this term is in the numerator, nothing happens until you get to ωo and then the response starts growing. While it if it is in the denominator nothing happens until you get to ωo and then things start falling. With this in mind, you can start at very low frequencies (like 0) and see what the gain is there. Then expect it to start increasing when you get to about 5 r/s, and then expect it to level off when you get to about 50 r/s. For very high frequencies, you should be able to see what the gain will be.