# May I please have help calculating the noise bandwidth for the circuit below

I am trying to calculate the noise bandwidth for the circuit below. This question and diagram is taken from the book "Low-noise electronic system design" by C.D. Motchenbacher and J.A. Connelly (if anyone knows of a solution manual for this book I would be eternally grateful).

To begin this problem I know that I need the transfer function (I hope I got this part right at least).

$$H(s)=\frac{100RCs}{(1+RCs)(1+RCs)}$$ After using the information from the circuit diagram to simplify the transfer function I have the equation: $$H(s)=\frac{100s}{(1+s)(1+s)}$$ To actually determine the noise bandwidth I need to use the formula:

$$\Delta f=\frac{1}{H_{max}^2}\int_{0}^{\infty}|H(f)|^2\;df$$

From what I read in the text book I understand that to get from a function of s to a function of f I should substitute some sort of pole but I am not sure what it should look like. The examples in the textbook are of simpler cases like first/second order low pass filters where using the relationship $RC=\frac{1}{2\pi f_c}$ and substituting a pole $s=j\omega$ results in a nice function of f, and determining the square magnitude of that transfer function is simple. However, that made sense to me because those filters have one corner frequency but a bandpass filter like the circuit below has two. Every substitution I can think of ends up with a really nasty function that I can't integrate. I know I am missing something fundamental but I just don't see it.

I guess what I would really like is guidance on how to determine |H(f)|^2 for circuits like this.

simulate this circuit – Schematic created using CircuitLab

Thank you in advance to anyone who is willing to assist me.

• why did you assume RC=1 in your simplification? – Tony Stewart Sunnyskyguy EE75 Apr 5 '17 at 3:26
• Noise -3dB bandwidth will be something near $1/2\pi$, But R must be specified for noise current. – Tony Stewart Sunnyskyguy EE75 Apr 5 '17 at 12:08
• The diagram and all the information on it is exactly as is in the text. I made no assumptions about the values of the components. – user144371 Apr 7 '17 at 0:10
• Your reduction for H(s) is only valid for RC=1 hence wrong – Tony Stewart Sunnyskyguy EE75 Apr 7 '17 at 2:51
• But RC=1s is given as part of the problem? why wouldn't I use that? – user144371 Apr 8 '17 at 12:51

A key thing to remember is that noise does not come in an amplitude of volts or amps, it comes in a noise power. For a resistor ( a white noise source) this noise is:

$$v_{n}^2 = \int_{0}^{\infty}S_v(f)df = 4kTR$$ the units are $\frac{V}{\sqrt{Hz}}$ or $\frac{V^2}{Hz}$

2) How to integrate

With noise its mostly better to think of it graphically. So this is a bandpass filter with a gain of 100. The bandpass filter is going to have a high pass, then a pass band and a low pass.

At the end of the day, if you were to have an actual noise input on Vin (like a white noise generator (a generator that has an equal amplitude of Gaussian noise on every frequency -- on average) or a resistor (thermal noise is the same thing as white noise), you would see band limited noise on a scope (with an FFT), the noise would start near zero, then go to 100x the amplitude as it approaches the first time constant of the band pass filter. Then stay at 100x the amplitude until the second time constant and then roll of back down near zero after the second time constant.

We can actually represent these as areas and do the integration without integrals and you can split them up. I will not show you all the math and deprive you of valuable learning.

Here is what the integration is for the passband section:

My $|H(f)|^2 = 100*Vin$ (and Vin should be in units of $\frac{V}{\sqrt{Hz}}$) and the area of integration is from $\tau_1$ to $\tau_2$ but we want this in frequency so we use the age old forumla $f_c = \frac{1}{2\pi\tau}$. Another gotcha is you haven't specified what noise source you are actually trying to integrate, I'm assuming Vin. Ether way this will give you the tools to find any noise source.

If we know that the amplitude is constant H(f)^2 becomes 100*Vin (or whatever your noise source is)

$$\int_{f_{c1}}^{f_{c2}}|H(f)|^2\;df = \int_{f_{c1}}^{f_{c2}}\;df* 100*Vin = \Delta f*100*Vin$$

This means that you can solve these geometrically (remember your in log land with a 20dB rolloff) and the other two are triangles.

However if you want to double check with an integral $|H(f)|^2 = \frac{100*Vin}{1+RCs}$ where $s= j\omega$ and (actually I'll finish this tommrow) $$\int_{0}^{f_{c1}}|H(f)|^2\;df$$

• This is a good analysis and shows unique values for T1 and T2 even though the question indicates RC values are all the same, thus T1,T2 neglects the situation where T1 =T2 with -3dB at the asymptote or breakpoint and 2nd order reduction of -6dB from 100xVin or 40dB to be 34dB at centre, which affects noise bandwidth points of half power to become gains at 34-3=31dB for f1 & f2 (-3dB) and graph is log H(f) vs log f. – Tony Stewart Sunnyskyguy EE75 Apr 7 '17 at 3:32
• For RC=1, therefore I compute half power signal BW at (-3dB) points at H(f)=31dB = 40-6-3 to be 67mHz and 377 mHz (that's millihertz) in this hypothetical question The centre frequency is of course f=1/2pi = 159 mHz for RC=1 – Tony Stewart Sunnyskyguy EE75 Apr 7 '17 at 3:37

Firstly, it seems that your computation of the transfer function H(s) is incorrect.

Let us find the voltage across the resistor R1. Let this voltage be Va.

simulate this circuit – Schematic created using CircuitLab

\begin{eqnarray*} \dfrac{V_{a}}{V_{in}} &=& \dfrac{R{1} || (R_{2}+\dfrac{1}{sC})}{(R{1} || (R_{2}+\dfrac{1}{sC}))+\dfrac{1}{sC}} \\ &=& \dfrac{s(1+s)}{s^2+3s+1} \end{eqnarray*}

wherein I assume that $$R_{1} = R_{2} = R = 1M \Omega$$ $$C=1 \mu F \hspace{2mm} such \hspace{2mm} that \hspace{2mm} RC=1$$

Now, Va gets amplified by a gain of 100, and then Vout occurs across the final capacitor.

\begin{eqnarray*} \dfrac{V_{out}}{V_{a}} &=& \dfrac{\dfrac{100}{sC}}{R_{2}+\dfrac{1}{sC}} \\ &=& \dfrac{100}{1+s} \end{eqnarray*}

\begin{eqnarray*} H(s) = \dfrac{V_{out}}{V_{in}} &=& \dfrac{100}{1+s} \times \dfrac{s(1+s)}{s^2+3s+1} &=& \dfrac{100s}{s^2+3s+1} \end{eqnarray*}

Moving on to the analysis of the noise bandwidth. As you mentioned, we make use of the fact that $$s = j \omega = j 2 \pi f$$. H(f) can now be expressed as:

\begin{eqnarray*} H(f) = \dfrac{j(200 \pi f)}{(1-4 \pi^2 f^2) +j (6 \pi f)} \end{eqnarray*}

Taking the magnitude of this complex ratio and then squaring it,

\begin{eqnarray*} |H(f)|^2 &=& \dfrac{(200 \pi f)^2}{(1-4 \pi^2 f^2)^2 +(6 \pi f)^2} &=& \dfrac{(200 \pi f)^2}{16 \pi^4 f^4 + 28 \pi^2 f^2 +1} \end{eqnarray*}

Finding Hmax can be done by simply equating the first derivative of H(f) to 0 and then solving for the value of f that yields Hmax. However, since this is rather laborious, I've taken help from Wolfram Alpha's computational knowledge engine to carry out the computation.

As for the part concerning the integral of |H(f)|^2, it can be done by partial fractions. Once again, Wolfram alpha tells us that 16 pi^4 f^4 + 28 pi^2 f^2 + 1 =0 can be factorized.

In conclusion, the integral can be computed on Wolfram Alpha as well to give a neat answer of 0.75!

Please do edit the post if there are blatant errors!

• In the square of the amplitude, isn't there a ^2 missing in the upstairs? – Timo Apr 5 '17 at 9:22
• Oh of course! How did I miss that. Let me make the necessary changes. – V-Red Apr 6 '17 at 3:25
• I don't understand how I can pull the amplifier out of the circuit and make it a passive network? I have many more questions, and I am trying to compile them all and some more of my thoughts into a response. Thank you for your help so far. – user144371 Apr 7 '17 at 0:08
• YOu cannot pullout the amp as it buffers then 2nd passive filter, otherwise the impedance loading of the 2nd filter affects the 1st such that it is only valid if R2 is >> R1 meaning the impedance of the 1st filter must be much lower than the 2nd to exclude an Amp – Tony Stewart Sunnyskyguy EE75 Apr 7 '17 at 2:54
• Although a nice analysis, it is not the same circuit without an Amp. In this passive schematic, the gain should be H(f) ~ -9.5dB midband due to attenuation when |Zc(f)|=R at the breakpoint. , if buffered in between it would be -6dB and with 40dB gain = 34dB , – Tony Stewart Sunnyskyguy EE75 Apr 7 '17 at 3:08