# Filtering periodic signals

 Page 3 / 3

## Low-pass filtering with sound

Use wavread to load the sound castanets44m.wav. Perform low-pass filtering with the filter defined above, starting with a = 500*2*pi , but also try different values.

Play the original and the low-passed version of the sound. Plot their frequency content (Fourier transforms) as well.

## Optional: low-pass filtering

1. Create an impulse train as the input signal x(t) using the following MATLAB command, >>x = repmat([zeros(1, 99) 1], 1, 5);
2. Use the low-pass filter defined earlier to low-pass the impulse train. Choose a cutoff of 20.
3. Plot the two signals x(t) and y(t) separately using the subplot command. These should be plotted versus the time vector. Label the axes and title each graph appropriately.
4. Look at the plots. Can you explain what is happening to the spike train?

## High-pass filtering

An ideal high-pass filter eliminates low frequency components entirely, as in: ${H}_{H}^{ideal}\left(\omega \right)=\left\{\begin{array}{cc}0& |\omega | A real high-pass filter typically has low but non-zero values for $|{H}_{L}\left(\omega \right)|$ at low frequencies, and a gradual (rather than an immediate) rise in magnitude as frequency increases. The simplest (and least effective) high-pass filter is given by (e.g. using an RC circuit): ${H}_{H}\left(\omega \right)=1-{H}_{L}\left(\omega \right)=1-\frac{\alpha }{\alpha +j\omega },\text{}\alpha =\text{cutoff frequency}\text{.}$ This filter can be implemented in the same way as the low pass filter above.

## High-pass filtering with sound

The high-pass filter can be implemented in MATLAB much the same way as the low-pass filter. Perform high-pass filtering with the filter defined above on the sound castanets44m.wav. Start with a = 2000*2*pi , but also try different values.

Play the original and the high-passed version of the sound. The filtered signal may be to be scaled so that both have the same range on the Y-axis. Plot their frequency responses as well.

## Sound filtering

• Kick'n Retro 235 Inc. recorded a session of a trumpet and drum kit together for their new release. The boss doesn't like the bass drum in the background and wants it out. Unfortunately, there was a malfunction in the mixing board and instead of having two separate tracks for the drums and the trumpet, the sounds mixed together in one track. In order to get this release out on time you will have to use some filtering to eliminate the bass drum from the sound. There is not enough time to bring the drummer and trumpet player back in the studio to rerecord the track.
• Click here to download the mixed.wav sound ( Fs = 8000 Hz). The mixed sound is created from bassdrum.wav, hatclosed.wav, and shake.mat.
• Try to do something easy but approximate first, and then, if you have more time, see how clean you can get the sound. You may find it helpful to look at the Fourier domain representation of the sounds, but you may not use the individual sounds in your solution.
• Now try to eliminate the trumpet sound, leaving only the drums left in the sound.
• Hint: use high-pass and low-pass filtering.
• Another hint: If you want a more powerful filter, you can try using multiple a/(a+jw) terms in series. Each extra term raises the order of the filter by one and higher order filters have a faster drop-off outside of their passing region.

## Bonus problem: sound filtering

• Imagine you recorded a trumpet and rainstick together, so that you have the signal, mixedsig = shake + 10*rainstick .
• It turns out the producer thinks the rainstick is too new-age and wants it out of the recording. Pretend you do not have the original signals shake or rainstick. Can you take the signal mixedsig and process it to get (approximately) only the trumpet sound (shake) out? Try to do something easy but approximate first, and then if you have more time, see how good a reproduction of shake you can get. You may find it helpful to look at Fourier domain of the sounds, but you may not use rainstick.mat or shake.mat in your solution.

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!