# Modulation and coding course- lecture 4

Chia sẻ: Trần Huệ Mẫn | Ngày: | Loại File: PDF | Số trang:23

0
60
lượt xem
10

## Modulation and coding course- lecture 4

Mô tả tài liệu

Receiver structure -Impact of AWGN and ISI on the transmitted signal -Optimum filter to maximize SNR -Matched filter receiver and Correlator receiver

Chủ đề:

Bình luận(0)

Lưu

## Nội dung Text: Modulation and coding course- lecture 4

1. Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 4
2. Last time we talked about: Receiver structure Impact of AWGN and ISI on the transmitted signal Optimum filter to maximize SNR Matched filter receiver and Correlator receiver Lecture 4 2
3. Receiver job Demodulation and sampling: Waveform recovery and preparing the received signal for detection: Improving the signal power to the noise power (SNR) using matched filter Reducing ISI using equalizer Sampling the recovered waveform Detection: Estimate the transmitted symbol based on the received sample Lecture 4 3
4. Receiver structure Step 1 – waveform to sample transformation Step 2 – decision making Demodulate & Sample Detect z (T ) Threshold ˆ mi r (t ) Frequency Receiving Equalizing comparison down-conversion filter filter For bandpass signals Compensation for channel induced ISI Received waveform Baseband pulse Baseband pulse Sample (possibly distored) (test statistic) Lecture 4 4
5. Implementation of matched filter receiver Bank of M matched filters z1 (T ) s (T − t ) ⎡ z1 ⎤ * 1 Matched filter output: r (t ) ⎢ M ⎥=z z Observation ⎢ ⎥ vector ⎢zM ⎥ ⎣ ⎦ sM (T − t ) * zM (T ) zi = r (t ) ∗ s ∗i (T − t ) i = 1,..., M z = ( z1 (T ), z 2 (T ),..., z M (T )) = ( z1 , z 2 ,..., z M ) Lecture 4 5
6. Implementation of correlator receiver Bank of M correlators s ∗1 (t ) T z1 (T ) ∫0 ⎡ z1 ⎤ Correlators output: r (t ) ⎢M⎥ =z z Observation ⎢ ⎥ ∗ vector s M (t ) T ⎢zM ⎥ ⎣ ⎦ ∫0 z M (T ) z = ( z1 (T ), z 2 (T ),..., z M (T )) = ( z1 , z 2 ,..., z M ) T zi = ∫ r (t )si (t )dt i = 1,..., M 0 Lecture 4 6
7. Today, we are going to talk about: Detection: Estimate the transmitted symbol based on the received sample Signal space used for detection Orthogonal N-dimensional space Signal to waveform transformation and vice versa Lecture 4 7
8. Signal space What is a signal space? Vector representations of signals in an N-dimensional orthogonal space Why do we need a signal space? It is a means to convert signals to vectors and vice versa. It is a means to calculate signals energy and Euclidean distances between signals. Why are we interested in Euclidean distances between signals? For detection purposes: The received signal is transformed to a received vectors. The signal which has the minimum distance to the received signal is estimated as the transmitted signal. Lecture 4 8
9. Schematic example of a signal space ψ 2 (t ) s1 = (a11 , a12 ) ψ 1 (t ) z = ( z1 , z 2 ) s 3 = (a31 , a32 ) s 2 = (a21 , a22 ) s1 (t ) = a11ψ 1 (t ) + a12ψ 2 (t ) ⇔ s1 = (a11 , a12 ) Transmitted signal s2 (t ) = a21ψ 1 (t ) + a22ψ 2 (t ) ⇔ s 2 = (a21 , a22 ) alternatives s3 (t ) = a31ψ 1 (t ) + a32ψ 2 (t ) ⇔ s 3 = (a31 , a32 ) Received signal at z (t ) = z1ψ 1 (t ) + z2ψ 2 (t ) ⇔ z = ( z1 , z 2 ) matched filter output Lecture 4 9
10. Signal space To form a signal space, first we need to know the inner product between two signals (functions): Inner (scalar) product: ∞ < x(t ), y (t ) >= ∫ −∞ x(t ) y * (t )dt = cross-correlation between x(t) and y(t) Properties of inner product: < ax (t ), y (t ) >= a < x(t ), y (t ) > < x(t ), ay (t ) >= a * < x(t ), y (t ) > < x(t ) + y (t ), z (t ) >=< x(t ), z (t ) > + < y (t ), z (t ) > Lecture 4 10
11. Signal space … The distance in signal space is measure by calculating the norm. What is norm? Norm of a signal: ∞ x(t ) = < x(t ), x(t ) > = ∫ x(t ) dt = E x 2 −∞ = “length” of x(t) ax(t ) = a x(t ) Norm between two signals: d x , y = x(t ) − y (t ) We refer to the norm between two signals as the Euclidean distance between two signals. Lecture 4 11
12. Example of distances in signal space ψ 2 (t ) s1 = (a11 , a12 ) E1 d s1 , z ψ 1 (t ) E3 z = ( z1 , z 2 ) d s3 , z E2 d s2 , z s 3 = (a31 , a32 ) s 2 = (a21 , a22 ) The Euclidean distance between signals z(t) and s(t): d si , z = si (t ) − z (t ) = (ai1 − z1 ) 2 + (ai 2 − z 2 ) 2 i = 1,2,3 Lecture 4 12
13. Orthogonal signal space N-dimensional orthogonal signal space N characterized by is N linearly independent functions { j (t )}j =1 called basis ψ functions. The basis functions must satisfy the orthogonality condition 0≤t ≤T T < ψ i (t ),ψ j (t ) >= ∫ψ i (t ) * (t )dt = K iδ ji ψj 0 j , i = 1,..., N where ⎧1 → i = j δ ij = ⎨ ⎩0 → i ≠ j If all K i = 1, the signal space is orthonormal. Lecture 4 13
14. Example of an orthonormal bases Example: 2-dimensional orthonormal signal space ⎧ 2 ⎪ψ 1 (t ) = cos(2πt / T ) 0≤t
15. Signal space … Any arbitrary finite set of waveforms {si (t )}iM1 = where each member of the set is of duration T, can be expressed as a linear combination of N orthonogal waveforms ψ { j (t )}Njwhere N ≤ M . =1 N si (t ) = ∑ aijψ j (t ) i = 1,..., M j =1 N≤M where j = 1,..., N T 1 1 aij = < si (t ),ψ j (t ) >= ∫ si (t )ψ j (t )dt 0≤t ≤T * Kj Kj 0 i = 1,..., M N Ei = ∑ K j aij 2 s i = (ai1 , ai 2 ,..., aiN ) j =1 Vector representation of waveform Waveform energy Lecture 4 15
16. Signal space … N si (t ) = ∑ aijψ j (t ) s i = (ai1 , ai 2 ,..., aiN ) j =1 Waveform to vector conversion Vector to waveform conversion ψ 1 (t ) ψ 1 (t ) T ai1 ∫ ai1 ⎡ai1 ⎤ ⎡ai1 ⎤ ⎢M⎥ 0 sm si (t ) ⎢ M ⎥ = sm si (t ) ψ N (t ) ⎢ ⎥ ⎢ ⎥ ψ N (t ) T ⎢aiN ⎥ ⎣ ⎦ ⎢aiN ⎥ ⎣ ⎦ ∫0 aiN aiN Lecture 4 16
17. Example of projecting signals to an orthonormal signal space ψ 2 (t ) s1 = (a11 , a12 ) ψ 1 (t ) s 3 = (a31 , a32 ) s 2 = (a21 , a22 ) s1 (t ) = a11ψ 1 (t ) + a12ψ 2 (t ) ⇔ s1 = (a11 , a12 ) Transmitted signal alternatives s2 (t ) = a21ψ 1 (t ) + a22ψ 2 (t ) ⇔ s 2 = (a21 , a22 ) s3 (t ) = a31ψ 1 (t ) + a32ψ 2 (t ) ⇔ s 3 = (a31 , a32 ) T aij = ∫ si (t )ψ j (t )dt j = 1,..., N i = 1,..., M 0≤t ≤T 0 Lecture 4 17
18. Signal space – cont’d To find an orthonormal basis functions for a given set of signals, Gram-Schmidt procedure can be used. Gram-Schmidt procedure: Given a signal set {si (t )}i =1 compute an orthonormal basis { j (t )}j =1 ψ M N , 1. Define ψ 1 (t ) = s1 (t ) / E1 = s1 (t ) / s1 (t ) i −1 2. For i = 2,..., M compute d i (t ) = si (t ) − ∑ < si (t ),ψ j (t ) > ψ j (t ) If d i (t ) ≠ 0 let ψ i (t ) = d i (t ) / d i (t ) j =1 If , (t = 0 d ido) not assign any basis function. 1. Renumber the basis functions such that basis is { 1 (t ),ψ 2 (t ),...,ψ N (t )} ψ This is only necessary if d i (t ) = 0 for any i in step 2. Note that N ≤ M Lecture 4 18
19. Example of Gram-Schmidt procedure Find the basis functions and plot the signal space for the following transmitted signals: s1 (t ) s2 (t ) A T 0 T t −A 0 T t T Using Gram-Schmidt procedure: T ψ 1 (t ) s1 (t ) = Aψ 1 (t ) 1 E1 = ∫0 s1 (t ) dt = A 2 2 1 s2 (t ) = − Aψ 1 (t ) ψ 1 (t ) = s1 (t ) / E1 = s1 (t ) / A T s1 = ( A) s 2 = (− A) T 2 < s2 (t ),ψ 1 (t ) >= ∫ s2 (t )ψ 1 (t )dt = − A 0 T t 0 s2 s1 d 2 (t ) = s2 (t ) − (− A)ψ 1 (t ) = 0 ψ 1 (t ) -A 0 A Lecture 4 19
20. Implementation of matched filter receiver Bank of N matched filters z1 ψ (T − t ) ∗ ⎡ z1 ⎤ 1 Observation vector r (t ) ⎢ ⎥=z z ⎢ ⎥ ψ ∗ N (T − t ) ⎢zN ⎥ ⎣ ⎦ zN N si (t ) = ∑ aijψ j (t ) i = 1,..., M j =1 z = ( z1 , z 2 ,..., z N ) N≤M z j = r (t ) ∗ψ j (T − t ) j = 1,..., N Lecture 4 20