# calculate frequency at -6 db

I have to do an exercise. Given this signal: $H(f)=\frac{\frac{T}{2}}{1+(\pi f\frac{T}{2})^2}$ calculate the frequency at $-6 db$. How can I resolve this exercise? I tried this: $\frac{\frac{T}{2}}{1+(\pi f\frac{T}{2})^2}=10^{-0.6}$ but I don't know if it is correct

• So, plot it and see if the result you're getting agrees with the plot... (though think carefully what -6dB means for the amplitude - assuming $H\left(f\right)$ is the amplitude). I assume that it's -6dB wrt the peak? Commented Jun 14, 2012 at 10:34
• Hint: To an engineer (as opposed to a calculator), $6$ dB and $-6$ dB mean ratios of $4$ and $0.25$, and it is highly unlikely that your instructor intended for you to use the exact value of $10^{-0.6}$. Also, decibels are intended to measure power ratios but sometimes are used for voltage ratios as well. Depending on what else is in the question that you are being asked, you might be being asked to find the value of $f$ that makes the ratio $\left|\frac{H(f)}{H(0)}\right|^2$ equal $0.25$ or makes $\left|\frac{H(f)}{H(0)}\right|$ equal $0.25$ (cf. @HenryGomersall's comment). Commented Jun 14, 2012 at 11:17
• Your signal (or transfer function) is clearly lowpass and monotonic. With -6 dB I guess your teacher means find the frequency where $\frac{H(f)}{H(0)} = \frac{1}{2}$, which means solve for $f$ in $\pi f \frac{T}{2} = 1$. Commented Jun 14, 2012 at 19:20
• Mazzy, if you will get off your duff and actually write out what $\left|\frac{H(f)}{H(0)}\right|$ is (or just $\frac{H(f)}{H(0)}$ as niaren suggests since $H(f)$ is a positive real-valued function in this case), you will see that the solution reduces to what niaren is telling you it is. Commented Jun 15, 2012 at 13:47
• Read the last line of my first comment up to the first occurrence of $0.25$. Then use your calculator to find $\sqrt{0.25}$ which should be equal to $\frac{H(f)}{H(0)}$, right? Commented Jun 16, 2012 at 0:36