At last Limit x tends to 0, (Log base2 (1+x))/x = (log e)/log 2 = 1.44 I did small error, in that I wrote 1/x in denominator, I will remake this video soon.
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A telephone line normally has a bandwidth is from 300 to 3300 Hz assigned for data communication. The SNR is usually 3162. What will be the capacity for this channel?
To calculate the capacity of a communication channel, we use the Shannon-Hartley theorem. The theorem states that the channel capacity (C) in bits per second (bps) is given by: C = B log_2(1 + SNR) where: - B is the bandwidth of the channel in hertz (Hz). - SNR is the signal-to-noise ratio, expressed as a power ratio (not in decibels). Given: - The bandwidth B is from 300 Hz to 3300 Hz, so ( B = 3300 - 300 = 3000 ) Hz. - The signal-to-noise ratio (SNR) is 3162. Let's substitute these values into the Shannon-Hartley theorem formula: C = 3000 log_2(1 + 3162) First, compute ( 1 + 3162 ): [ 1 + 3162 = 3163 ] Next, we need to calculate ( log_2(3163) ). This can be done by converting the logarithm base 10 to base 2 using the change of base formula: [ log_2(3163) = log_{10}(3163)}/{log_{10}(2)}] Using a calculator, we find: [log_{10}(3163) = 3.5004] [log_{10}(2) = 0.3010] Now, divide these values: [ log_2(3163) = {3.5004}/{0.3010} =11.63] Finally, calculate the capacity: [ C = 3000 * 11.63 = 34890 bps ] Thus, the capacity of the telephone line channel is approximately 34,890 bits per second (bps).
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2:34 doesn't explain why the number of levels that can be separated without error is what it is. I think it involves some sort of coding with large hamming distances concept.
Your positive comments motivates me. Teachers like me just wants positive comment from student. Love from you guys means a lot to me. My goal is to create largest community of engineers in entire globe. Please help me by sharing this playlist with your friends.
(Noise + Signal)/Noise = m m should be as high as possible. If m is high then we can have more number of levels in qauntized signal. Practically it shows levels can be detected without error. By writing here it is difficult to explain but if you draw it then you will get it exactly.
@@EngineeringFunda Equation is of the form :- log₂ e(1 + x) / (x), in this case. But you have considered the formula for log₂ e(1 + x) / (1 / x). Please check.
eng mat bola kariye sir...na ap atche se samjha pate hay..na hume samajh ata he..adha adhura reh jata he..ya to apke khud k concept clear nhi hay .ya fir eng kamjor hay
Savdhan this channel is for ece ,cse wale durr rahe ye sir cse walon ke liye nhi hai we don't know abc of ece and he assumes we know everything and yha pr time pass ke liye aayen hai 8:32
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At last Limit x tends to 0,
(Log base2 (1+x))/x = (log e)/log 2 = 1.44
I did small error, in that I wrote 1/x in denominator, I will remake this video soon.
Ok thanks you sir 😊 your lectures are good sir
i guess it's quite randomly asking but do anyone know a good website to watch newly released series online?
@Emilio Reign Flixportal :P
@Rodrigo Darian thanks, I went there and it seems to work :D I appreciate it !!
@Emilio Reign No problem :)
S Latha is correct, it should be lim x -> 0 of log2(1 + x)/x = log2(e)
One can understand in one lecture .... Thank u sir
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I gave the 200th Like. 😎 Awesome and quick explanation.
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A telephone line normally has a bandwidth is
from 300 to 3300 Hz assigned for data
communication. The SNR is usually 3162.
What will be the capacity for this channel?
To calculate the capacity of a communication channel, we use the Shannon-Hartley theorem. The theorem states that the channel capacity (C) in bits per second (bps) is given by:
C = B log_2(1 + SNR)
where:
- B is the bandwidth of the channel in hertz (Hz).
- SNR is the signal-to-noise ratio, expressed as a power ratio (not in decibels).
Given:
- The bandwidth B is from 300 Hz to 3300 Hz, so ( B = 3300 - 300 = 3000 ) Hz.
- The signal-to-noise ratio (SNR) is 3162.
Let's substitute these values into the Shannon-Hartley theorem formula:
C = 3000 log_2(1 + 3162)
First, compute ( 1 + 3162 ):
[ 1 + 3162 = 3163 ]
Next, we need to calculate ( log_2(3163) ). This can be done by converting the logarithm base 10 to base 2 using the change of base formula:
[ log_2(3163) = log_{10}(3163)}/{log_{10}(2)}]
Using a calculator, we find:
[log_{10}(3163) = 3.5004]
[log_{10}(2) = 0.3010]
Now, divide these values:
[ log_2(3163) = {3.5004}/{0.3010} =11.63]
Finally, calculate the capacity:
[ C = 3000 * 11.63 = 34890 bps ]
Thus, the capacity of the telephone line channel is approximately 34,890 bits per second (bps).
@@EngineeringFunda thanks
Thank you sir for this..
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Is there any other method to prove this theorem using probability Dencity function
Sir i am not able to find source coding theorem in your channel. Could you please share a link if topic on this video is available in ur playlist.
Thanku sir for great teaching it is understandable good sir ❤️
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Sure sir ♥️✨
Osom lecture sir thank you 😊
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you are boss in control system.your all tutorial are helpful.please give more vedioes
2:34 doesn't explain why the number of levels that can be separated without error is what it is. I think it involves some sort of coding with large hamming distances concept.
sir aapne jo bhe padhya kuch smjh nhi ayaa , Thankyou
From where the noise came in the ofc cables?
AGWN concept is not used in this proof....?
Please explain the implicants of Shannon Hartley theorem
sir i have a math..if bandwidth (300 - 3200), and SNR 20 dB,what is the chAnnel capacity and,here which bandwidth
Bandwidth is higher cutoff - lower cutoff frequency
Lim x tends to zero (1+x)^(1/x)=e
Sir please provide the PDF of information theory.
sir what is n (eeta)?
Efficiency
Thanku sir❤️
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Thanks I understand a bit
Good pun
By the way we wont say forty four ....after point
point four four :X
Sir I didn't understand why u take the ration for calculating no of levels of un effected by noise
It is ratio of voltages with noise signal and noise
But why u have taken like that sir .....
Means I just want to know how that formula came
Thanks for your fast reply... For previous comment
(Noise + Signal)/Noise = m
m should be as high as possible.
If m is high then we can have more number of levels in qauntized signal. Practically it shows levels can be detected without error. By writing here it is difficult to explain but if you draw it then you will get it exactly.
This is how fast God can think.Assuming God is omnipotent.
Yes, GOD is omnipotent that I feel.
Are diwane God mind se think karega shannon Hartley theorem se nahi
Sir do channel capacity of Gaussian channel and channel capacity by Shannon same sir ?? Reply fast sir tmr i have external exam
What happened ur exam bro
Sir, i think some mistake in log2e's formula...
It's correct
@@EngineeringFunda how did u applied the formula of log e? I'm also confused here.
@@EngineeringFunda Equation is of the form :- log₂ e(1 + x) / (x), in this case. But you have considered the formula for log₂ e(1 + x) / (1 / x). Please check.
S/nb = x,
Then 1/x= nb/s but sir you used above expression in that expression of limit.
eng mat bola kariye sir...na ap atche se samjha pate hay..na hume samajh ata he..adha adhura reh jata he..ya to apke khud k concept clear nhi hay
.ya fir eng kamjor hay
U r wrong with limit .......
Savdhan this channel is for ece ,cse wale durr rahe ye sir cse walon ke liye nhi hai we don't know abc of ece and he assumes we know everything and yha pr time pass ke liye aayen hai 8:32
👍
Thanks and welcome
Please say me sir
Clear
sir plz improve your teaching skill first then teach
Ok
Why u being so rude? U could have just said that u didn't grasp the concept. Or watched another video.
If u want to see then see else leave
sir i am them sure, that u don't know how great tutor u are 🫡🫡
Thank you so much 😀