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Quadrature modulation

In AM transmission where the baseband signal and its modulated passband version are real valued,the spectrum of the modulated signal has twice the bandwidth of the baseband signal.As pictured in [link] , the spectrum of the baseband signal is nonzero only forfrequencies between - B and B . After modulation, the spectrum is nonzero in the interval [ - f c - B , - f c + B ] and in the interval [ f c - B , f c + B ] . Thus the total width of frequencies occupied by the modulated signal is twicethat occupied by the baseband signal. This represents a kind of inefficiency or redundancyin the transmission. Quadrature modulation provides one way of removing this redundancyby sending two messages in the frequency ranges between [ - f c - B , - f c + B ] and [ f c - B , f c + B ] , thus utilizing the spectrum more efficiently.

To see how this can work, suppose that there are two message streams m 1 ( t ) and m 2 ( t ) . Modulate one message with a cosine to create the in-phase signal, and the other with (the negative of) a sine to form the quadrature signal. These are summed These are also sometimes modeled as the “real” and the “imaginary”parts of a single “complex valued” signal. This complex representation is explored more fully inAppendix [link] . to form

v ( t ) = m 1 ( t ) cos ( 2 π f c t ) - m 2 ( t ) sin ( 2 π f c t )

which is then transmitted. A receiver structure that can recover the two messagesis shown in [link] . The signal s 1 ( t ) at the output of the receiver is intended to recover the first message m 1 ( t ) . Similarly, the signal s 2 ( t ) at the output of the receiver is intended to recover the (negative of the) second message m 2 ( t ) .

To examine the recovered signals s 1 ( t ) and s 2 ( t ) in [link] , first evaluate the signals before the lowpass filtering. Using the trigonometric identities [link] and [link] , x 1 ( t ) becomes

x 1 ( t ) = v ( t ) cos ( 2 π f c t ) = m 1 ( t ) cos 2 ( 2 π f c t ) - m 2 ( t ) sin ( 2 π f c t ) cos ( 2 π f c t ) = m 1 ( t ) 2 ( 1 + cos ( 4 π f c t ) ) - m 2 ( t ) 2 ( sin ( 4 π f c t ) ) .
In a quadrature modulation system, two messages m_1(t) and m_2(t) are modulated by two sinusoids of the same frequency, sin(2πf_ct) and cos(2πf_ct). The receiver then demodulates twice and recovers the original messages after lowpass filtering.
In a quadrature modulation system, two messages m 1 ( t ) and m 2 ( t ) are modulated by two sinusoids of the same frequency, sin ( 2 π f c t ) and cos ( 2 π f c t ) . The receiver then demodulates twiceand recovers the original messages after lowpass filtering.

Lowpass filtering x 1 ( t ) produces

s 1 ( t ) = m 1 ( t ) 2 .

Similarly, x 2 ( t ) can be rewritten using [link] and [link]

x 2 ( t ) = v ( t ) sin ( 2 π f c t ) = m 1 ( t ) cos ( 2 π f c t ) sin ( 2 π f c t ) - m 2 ( t ) sin 2 ( 2 π f c t ) = m 1 ( t ) 2 sin ( 4 π f c t ) - m 2 ( t ) 2 ( 1 - cos ( 4 π f c t ) ) ,

and lowpass filtering x 2 ( t ) produces

s 2 ( t ) = - m 2 ( t ) 2 .

Thus, in the ideal situation in which the phases and frequencies of the modulation and the demodulation are identical, bothmessages can be recovered. But if the frequencies and/orphases are not exact, then problems analogous to those encountered with AM will occur in the quadrature modulation.For instance, if the phase of (say) the demodulator x 1 ( t ) is not correct, then there will be some distortion or attenuation in s 1 ( t ) . However, problems in the demodulation of s 1 ( t ) may also cause problems in the demodulation of s 2 ( t ) . This is called cross-interference between the two messages.

Use AM.m as a starting point to create a quadrature modulation system that implements the block diagram of [link] .

  1. Examine the effect of a phase offset in the demodulating sinusoids of the receiver, so that x 1 ( t ) = v ( t ) cos ( 2 π f c t + Φ ) and x 2 ( t ) = v ( t ) sin ( 2 π f c t + Φ ) for a variety of Φ . Refer to Exercise  [link] .
  2. Examine the effect of a frequency offset in the demodulating sinusoids of the receiver, so that x 1 ( t ) = v ( t ) cos ( 2 π ( f c + γ ) t ) and x 2 ( t ) = v ( t ) sin ( 2 π ( f c + γ ) t ) for a variety of γ . Refer to Exercise  [link] .
  3. Confirm that a ± 1 phase error in the receiver oscillator corresponds to more than 1% cross-interference.

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Source:  OpenStax, Software receiver design. OpenStax CNX. Aug 13, 2013 Download for free at http://cnx.org/content/col11510/1.3
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