Suppose the signal given is $x(t)=(cos(5t)+e^{-2t})u(t)$ and the goal is to find the bandwidth of this signal. We know the bandwidth, $B$, is just $f_{max}-f_{min}$.

Our first step is taking the Laplace Transform of $x(t)$,


which results in


At this point, we need to find the magnitude of our Laplace Transform, $X(s)$, and from what I understand this is easiest by changing $s$ to $j\omega$ by using $s=\sigma+j\omega$ with $\sigma=0$ in this case.

Leading us to


How would I go about finding the magnitude correctly in this case to find the bandwidth?


By combining ratios,


which leads us to


which should be


Using this, if I set $\omega=0$, I get $\frac{1}{2}$ and if I put any $\omega$ value of $2$ or more, it becomes a complex number.


Setting the denominator correctly for $j$,


At this point I feel like there is a mistake somewhere or I am going down the wrong path because this problem feels like more work that required since it literally just asks for

Find the bandwidth of the signal given by x(t)=(cos(5t)+e^(-2t))u(t)

  • $\begingroup$ The question would make more sense to me if it was convolution and not addition-- meaning a 5 rad/sec cosine going through a filter with an impulse response given by $e^{-2t}$. Are you certain there isn't a typo at that + sign? $\endgroup$ Commented Apr 21, 2022 at 9:51

1 Answer 1


Good work thus far.

First combine the two ratios with a common denominator using:

$$\frac{a}{b} + \frac{c}{d}= \frac{ad + bc}{bd}$$

Then you can determine the magnitude using the relationship for the magnitude of a complex number given as $a+jb$ is $\sqrt{a^2+b^2}$. This should clear up your remaining confusion.

The "bandwidth" then must be defined. Typically a "3 dB bandwidth" is used which refers to the half power points in the waveform (from fmin to fmax). This would be where the magnitude would be $1/\sqrt{2}$. (That choice is somewhat arbitrary, but since the frequency content extends to infinity, and getting infinitely small as it does, we need to define some sort of "cutoff". I will assume here that it is specifically when the magnitude reaches $1/\sqrt{2}$.

Additional comments on efforts made:

The OP has successfully combined the ratios, but the next step in computing the magnitude is in error. The approach is to group all the reals together noting that $\omega$ is real to complete the magnitude of a general complex number with real and imaginary components $a+jb$ to be $\sqrt{a^2+b^2}$. Note in that form that $a$ is the real component and $b$ is the imaginary component. For example, the numerator given correctly as $25+2j\omega -2\omega^2$ has real component $25-2\omega^2$ and imaginary component $2\omega$. Use these as the $a$ and $b$ in the expression I give to compute the magnitude (This is simply the cartesian distance on the complex plane). Similarly the denominator first needs to be completely multiplied out, and then group the imaginary and real components together and compute the magnitude for that.

Additional comments:

I concur with the OP's conclusion that the path I suggested above (which would then be easy to compute with MATLAB or Python if available) is quite cumbersome. Given we have the sum of the two functions, we can from superposition treat each one independently and from that make a reasonable estimate as to the maximum and minimum frequency- The OP has done this already and just needs to determine for each what the -3 dB frequency corner would be (for a 3 dB BW case, which is $1/\sqrt{2}$ in magnitude.

So given the composite function as:

$$X(\omega) = X_1(\omega) + X_2(\omega)$$

We have for $X_1(\omega)$:

$$X_1(\omega) = \frac{j\omega}{25-\omega^2}$$

$$|X_1(\omega)| = \bigg|\frac{\omega}{25-\omega^2}\bigg| \tag{1}\label{1}$$

The criteria for bandwidth was not specified, but if we state that it is when the magnitude reaches $1/\sqrt{2}$ (not relative to the max in this case since the maximum goes to infinity!), then we could solve for $\omega$ as:

$$ \bigg|\frac{\omega}{25-\omega^2}\bigg| = 1/\sqrt{2}$$

For this, the highest frequency is solved to be:

$$\omega_{hi} = \frac{1+\sqrt{51}}{\sqrt{2}} \approx 5.76$$

Note that a cosine on its own would have no bandwidth and would occupy the frequency of 5 rad/sec. However this is a cosine multiplied by a unit step function, which serves to extend it's bandwidth out to infinity going down by 1/f. The 5.76 radians/sec point is where the magnitude in frequency has dropped to $1/\sqrt{2}$.

And for $X_2(\omega)$:

$$X_2(\omega) = \frac{1}{j\omega+2}$$ $$|X_2(\omega)| = \frac{1}{\sqrt{\omega^2+4}}\tag{2}\label{2}$$

This has a frequency that extends to DC (the magnitude at DC is 0.5), and thus the lowest frequency = 0. Note that at the high frequency determined from the cosine of $\omega= 5.76$, $|X_2(\omega)| = 0.164$. To be accurate in determining the frequency for a specific magnitude such as $1/\sqrt{2}$, we would have to account for the magnitude from both $|X_1(\omega)|$ and $|X_2(\omega)|$ in which case I can't imagine the requested effort is to solve such a formula by hand. Below I have plotted the magnitude of the composite function using Matlab where the magnitude is given in dB, and the $1/\sqrt{2}$ levels would correspond to -3 dB, where we see that $f_{min} = 0$ and $f_{max}$ at -3dB is 5.98 Hz (and not depicted, but the magnitude at 5 rad/sec does go to infinity, I just didn't include that exact point when creating the plot):

magnitude vs frequency

Zoom in:

zoom in

The same plot on a log scale may provide further insight into what "bandwidth" might mean:

log plot

  • $\begingroup$ How is the jw in the numerator dealt with in regard to this problem? I'm confused if I'm supposed to take the square root and square of just the denominator or the numerator also. If thats is the case, I end up with sqrt(jw)/(sqrt(625-w^2)+1/5, is this correct? $\endgroup$ Commented Apr 20, 2022 at 23:02
  • $\begingroup$ for jw the w is the b I use in my example. First combine the two terms under a common denominator so that you have a single ratio of a numerator and denominator. I clarified this with an edit above. Is that clearer now? $\endgroup$ Commented Apr 20, 2022 at 23:26
  • $\begingroup$ I ended up with sqrt(625-8w^2)/sqrt((4-w^2)(625-w^4)), which at w=0 is 1/2. Due to dividing by zero at w=2, is the bandwidth just 2? I'm ignoring negative values with the 2 for bandwidth. $\endgroup$ Commented Apr 21, 2022 at 0:01
  • $\begingroup$ Add the rest of your work to the bottom of your question by editing it and I will check it... $\endgroup$ Commented Apr 21, 2022 at 0:36
  • $\begingroup$ @Dan Bosschen. My compliments to you for being so helpful to the denizens of this web site. $\endgroup$ Commented Apr 21, 2022 at 0:38

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