Noise reduction and square wave/sine function question

[niteshadw]

I have two simple questions but I'm not 100% on how to get the correct result.

1. "The noise is reduced by 6 dB" means its amplitude is cut to _%?
How is it calculated to 50%? I try 20log(Av)...

2. Does this wave contain 30k hertz sine function? (see attachment)
The answer is yes, but I try f=1/T, in this case T = 0.1 ms, so f is 10k, how is it 30k?

Thank you

 

[Swapnil]

1. "The noise is reduced by 6 dB" means its amplitude is cut to _%?
How is it calculated to 50%? I try 20log(Av)...

Yes that is what you should try.
20 \log(A_v) = -6 \Rightarrow A_v = 10^\frac{-6}{20} \approx .50

2. Does this wave contain 30k hertz sine function? (see attachment) The answer is yes, but I try f=1/T, in this case T = 0.1 ms, so f is 10k, how is it 30k?

You are right in saying that the frequency of your wave is 10kHz. However, I don't think that's what the question is asking. It is asking does this wave *contain* 30KHz sine function. That is, it asking if the Fourier series of this wave includes a sine term that has the frequency 30kHz = 3(10kHz)? The answer is yes. Here is why: The given function is odd which means that only the coefficients b_k's are non-zero. The function is also half-wave symmetric which means that only the odd harmonics of b_k are non-zero. Thus, your Fourier series should have the form:
f(t) = \sum_{k=1,3,5...}^{\infty} b_k \sin(k(10k)t)

Notice that one of the terms in the series is a sine term which has a frequency of 30kHz.

 

[niteshadw]

Thank you very much, that was much of a help! =)

Yes that is what you should try.
20 \log(A_v) = -6 \Rightarrow A_v = 10^\frac{-6}{20} \approx .50


You are right in saying that the frequency of your wave is 10kHz. However, I don't think that's what the question is asking. It is asking does this wave *contain* 30KHz sine function. That is, it asking if the Fourier series of this wave includes a sine term that has the frequency 30kHz = 3(10kHz)? The answer is yes. Here is why: The given function is odd which means that only the coefficients b_k's are non-zero. The function is also half-wave symmetric which means that only the odd harmonics of b_k are non-zero. Thus, your Fourier series should have the form:
f(t) = \sum_{k=1,3,5...}^{\infty} b_k \sin(k(10k)t)

Notice that one of the terms in the series is a sine term which has a frequency of 30kHz.

 

[berkeman]

1. "The noise is reduced by 6 dB" means its amplitude is cut to _%?
How is it calculated to 50%? I try 20log(Av)...

To me, this question is ambiguous. Noise voltage decreases by 6dB when cut in half. But noise power decreases by 3dB when cut in half. The problem hopefully was more explicit, or was stated in a context that implied either noise voltage (or current) or noise power.