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How many signal levels are required when the bandwidth is 2MHz and signal to noise ratio is 8 dB?
Wiki User
∙ 14y ago
Use Nyquist and Shannon Heartly theorem to solve this
Nyquist theorem says that
Channel Capacity C = 2 * Bandwidth * log2 (Number of Signal levels)
Shannon Heartly theorem says that
Channel Capacity C = Bandwidth * log2( 1 + SNR)
Important points to consider while solving
Bandwidth is expressed in Hz
SNR is expressed in dB it must be converted using dB value = 10 log10(SNR)
(10 dB = 10, 20 dB = 100, 30 dB = 1000 etc..)
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∙ 14y ago
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Signal to noise ratio?
Noise signal is any signal which interferes with the main signal and does not give any important information.Signal should always be twice to that of noise.
10 example of music and noise?
If you have an amplifier running but no signal coming in, you will hear a low hissing sound which you can make louder by turning up the volume. This is an example of noise. What differentiates noise from other sounds is the human mind. Signal is sound you want; noise is sound you don't want. You can also say signal carries information but noise is a signal which carries none. However to understand the information the thinking mind is indispensible! Much of noise is of a random nature, unlike useful signal which has a structure. Noise can (to a significant degree) therefore be separated from desired signal by mathematical methods.
What is the purpose of white noise?
White noise sounds like a hiss. It can be used in the sythesis of musical instruments or sound effects. It is random noise and can be used for signal analysis.
Is it possible to transmit a digital signal coded as square wave as used inside a computer using radio transmission without loss?
Digital signals are regularly transmitted by radio systems. The traditional way of doing it was by frequency-shift keying (FSK) where the frequency was changed slightly between two states which represented logical 0 and 1. Another method uses phase-shift keying (PSK) which uses two phase states of the carrier wave to represent 0 and 1. The main issue in transmitting data by radio is the bandwidth, and with 2-state modulation like FSK or PSK the bandwidth must exceed the bit rate. Another important issue is signal-to-noise ratio because radio signals can become very weak over long distances. Alternative systems are now available in which the modulation can have more than 2 states. For example four alternative phase states instead of two can be used for each symbol, which allows 4 bits to be transmitted per symbol instead of two, so the data rate is doubled for the same bandwidth. General quadature-amplitude modulation (QAM) keying employs M amplitude states and N phase states to transmit MN bits in each symbol. This might seem a good idea but the disadvantage with using high values of M and/or N is that it requires a very high signal-to-noise ratio. A popular choice of modulation for data transmission using radio links in which the bandwidth and the signal strength are limited is quadrature phase-shift keying (QPSK) which provides a good compromise between data rate, bandwidth and signal-to-noise ratio.
What is white noise?
White noise is random sound that contains an equal amount of all the frequency components across the spectrum one is testing. It sounds sorta like the hiss one hears when a television is tuned to a blank channel (if the television doesn't automatically mute the sound when no signal is present, however). If one has white noise and examines the signal level of, say, the 55 cycle per second signal component, it will be the same as, say, the 927 cycle per second signal component, or the 2651 cycle per second signal component. Check this out: http://en.wikipedia.org/wiki/White_noise
What is the relation of signal to noise ratio and signal bandwidth?
C=blog(1+s/n)
Which factor affect data rate of a link?
the bandwidth and the signal to noise ratio
Advantages of large bandwidth?
A very usefull advantage is the exchange of SNR(signal to noise ratio) with Bandwidth... as on increasing the bandwidth the power required for transmission get reduced to a great extent.. is given by the formula SNR2 ~ (SNR1) B1/B2 AS we can see on increasing the bandwidth the SNR is reduced greatly
What is the connection between the bandwidth of a signal and the data rate that it represents?
The data rate (C) is equal to the bandwidth (B) times the logarithm base 2 of 1 plus the signal-to-noise ratio (S/N) (how much interference is introduced in the transmission of data)C = B x log2(1 + S/N)So your data rate is directly proportional to your bandwidth. If you increase your bandwidth, your data rate will also increase provided the signal-to-noise ratio isn't affected.
What is shanon capacity?
digital bandwidth = analogue bandwidth * log2 (1+ SNR) where SNR = strenthe of signal power/ strength of noise larger the SNR it is better.
Why high frequency signal is more affected by noise compare to low frquency signal?
AS FREQUENCY INCREASE THE BANDWIDTH INCREASE. AS WE KNOW NOISE HAVE LARGER BANDWIDTH. SO ITS AFFECT HIGH FREQUENCY SIGNAL. BUT LOW FREQUENCY SIGNAL HAVE LOW BANDWIDTH SO IT IS LESS AFFECTED BY NOISE. ALSO WE KNOW QUALITY FACTOR= CUTOFF FREQUENCY / BANDWIDTH. SO AS FREQUENCY INCREASE B.W. INCREASE SO QUALITY DEGRADE. CUTOFF FREQUNCY AND THE TERM FREQUENCY (USED HERE) IS DIFFERENT. CUTTOFF FREQUNCY IS USED IN FILTER. PRABIR KUMAR SETHY prabirsethy.05@gmail.com
What is the maximum data rate sustainable by a noisy channel with a bandwidth of 10KHz and a signal to noise ratio of 2047?
110kbps
What are the three interrelated major elements of cable bandwidth?
Distance, Frequency, and Signal-level-to-noise-level ratio (SNR)
What is difference between amplitude modulation and frequency modulation with respect to signal to noise ratio?
In FM noise is low as compared to AM. The AM signal covers more distance than FM signal that's why it gets more distorted.How ever the information in AM signal does not lose but the noise effects it more than that of FM signal,so we get signal with more noise than that of FM signal For FM Carson's rule Main article: Carson bandwidth rule A rule of thumb, Carson's rule states that nearly all (~98%) of the power of a frequency-modulated signal lies within a bandwidth of where , as defined above, is the peak deviation of the instantaneous frequency from the center carrier frequency . Noise quieting The noise power decreases as the signal power increases; therefore the SNR goes up significantly. Thanks, By tauseef ahmed
What is the difference between nyquist and shannon limit?
nyquist limit is for noiseless channel and given by 2Hlog2V where V are the discrete levels and H is the Bandwidth while, shanon limit is for channel with noise and given by H log2(1+S/N) where S is the signal power and N is the noise power.
What are the disadvantages of delta modulation?
1. The bandwidth for transmission of PAM signalis very very large compared to its maximum frequency content. 2. The Amplitude of PAM Pulses varies accordingly to modulated signal. Therefore interference of noise is maximum for the PAM signal and this noise cannot be removed very easily. 3. Since amplitude of PAM signal varies, this also varies the peak power required by the transmitter with modulating signal.
What is the effect of bandwidth on a digital signals?
Bandwidth is an inherent characteristic of a given transmission channel, or is determined by the narrowest-bandwidth component of the system. The bandwidth of a channel will limit the possible attainable data rates. This is shown simply by Shannon's Theorem, which states C = B log_2 (1+P_signal/P_noise), where C is the channel capacity in bps, B is the channel bandwidth in Hz, P_signal is the power of the detected signal in W, and P_noise is the noise power of the detected signal in W. As an example, consider a standard phone line (i.e., using a dial-up modem). Standard phone lines have a bandwidth of about 3.4 kHz and a signal-to-noise ratio of about 10,000. Using this information, we get C = (3.4 kHz) log_2 (1+10,000) = 45 kbps. Dial-up modems can actually get as high as 56 kbps, but that is beyond the scope of this question. In general Shannon's Theorem can provide a fairly accurate way to predict the possible data rates for a given transmission channel if the bandwidth and resulting signal and noise powers are known.
