# Bandwidth of the system

1. Dec 11, 2006

### LM741

hey all.

i read that the definition of the bandwidth of a system is the frequency range up until the signal's power (at DC) drops by -3dB.

This obviously only applies to a first order system , right?

Surely for a second order - it is defined as the range of frequency up until the power drops by -6dB?

thanks

2. Dec 11, 2006

### Staff: Mentor

Your first statement applies to a low-pass function. If the transfer function is a bandpass, then the bandwidth is generally measured to the 3dB points on either side of the passband. It doesn't matter what order the system is, you usually use the -3dB points as the shoulders.

3. Dec 11, 2006

### Staff: Mentor

4. Dec 11, 2006

### LM741

Thanks - but the reason why asked the above is because i am given the following system: H(s) = 1/ (s+4)^2, and asked to find the bandwidth of the system.
It can't be w=4 (If we wish to conform to the definition of bandwidth), because at this point we have a -6dB power drop.
On the other hand, if i was given the system as: H(s) = 1/(s+4), then the bandwidth would be equal to 4, i.e w=4

thanks again

5. Dec 11, 2006

### Staff: Mentor

Is your H(s) a power or voltage transfer function? Remember that -3dB is not the 1/2 signal point, it's a 1/2 power point. The signal at -3dB is $$\frac{1}{\sqrt2}$$

6. Dec 11, 2006

### LM741

power = signal drops by half DC value (or DC power??).
voltage or current = signal drops to 70 percent of DC value.
They are still both regarded as -3dB points by applying the corresponding equation:
for power : 10 log(P/2)
for voltage or current :20 log(V/srt(2)).

thanks

It can be seen as a voltage ...but i don't think in this case it will make a difference...

7. Dec 12, 2006

### LM741

any sugestions guys...?

8. Dec 13, 2006

### Corneo

I would solve for the value of $s_0$ such that $H(s_0) = \frac {1}{\sqrt 2} H_{max}$. Since the max is clearly 1, just solve for the denominator of H(s) $$(s+4)^2 = \sqrt {2}$$

9. Dec 15, 2006

### LM741

thanks - also thought about doing it that way and sticking to the definition.