?? 11.htm
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<P><IMG height=42 src="11.files/5.3.ht18.gif" width=448 border=0></P>
<P>從而可知</P>
<P><IMG height=45 src="11.files/5.3.ht19.gif" width=323 border=0></P>
<P>最后有</P>
<P>[<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">Δ</SPAN>W(k)]<SUP>2</SUP>-[<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">Δ</SPAN>W(k-1)]<SUP>2</SUP><0
(2.68)</P>
<P>式(2．68)說明引理的性質(zhì)(1)成立。</P>
<P>根據(jù)性質(zhì)(1)，則當(dāng)k——<SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">∞</SPAN>，則有w(k)＝W<SUB>0</SUB>．故而在式(2．66)中兩邊都為0。這也就是必定有</P>
<DIV align=center>
<CENTER>
<TABLE cellSpacing=0 cellPadding=0 width="80%" border=0>
<TBODY>
<TR>
<TD width="50%"><IMG height=51 src="11.files/5.3.ht20.gif" width=288
border=0></TD>
<TD width="50%">(2.69)</TD></TR></TBODY></TABLE></CENTER></DIV></TD></TR>
<TR>
<TD width="100%" height=414>
<P>可見，引理的性質(zhì)(2)成立。 </P>
<P>證畢。</P>
<P>有了上面的引理，就可以給出由式(2．53)—(2．59)組成的控制結(jié)構(gòu)對對象式(2．52)執(zhí)行適應(yīng)控制的閉環(huán)性質(zhì)定理。</P>
<P>定理：在對象由式(2．52)描述的控制中，式(2．53)—(2．59)構(gòu)成的適應(yīng)控制有如下的閉環(huán)性質(zhì)：</P>
<P>(1)輸入信號u(t)，輸出信號y(t)都是有界的。</P>
<P><IMG height=30 src="11.files/5.3.ht21.gif" width=230 border=0></P>
<P>證明：</P>
<P>設(shè)系統(tǒng)的跟蹤誤差用e'(k)表示</P>
<P>e'(k)=y(k)-r(k)
(2.70)</P>
<P>y(k)由式(2．61)給出。</P>
<P>r(k)可由式(2．57)，(2．58)，(2．59)求出，先用W<SUB>n+1</SUB>(k)乘〔2．59)兩邊，則有</P>
<P>W<SUB>n+1</SUB>(k-1)u(k-1)=r(k)+W<SUB>1</SUB>(k-1)(-X<SUB>1</SUB>(k-1))+.....,+W<SUB>n</SUB>(k-1)(-X<SUB>n</SUB>(k-1))+W<SUB>n+2</SUB>(k-1)(-X<SUB>n+2</SUB>(k-1))+......,Wn+m(k-1)(-Xn+m(k-1))</P>
<P>整理后有</P>
<P>r(k)=W<SUB>1</SUB>(k-1)X<SUB>1</SUB>(k-1)+......,+W<SUB>n</SUB>(k-1)X<SUB>n</SUB>(k-1)+W<SUB>n+1</SUB>(k-1)u(k-1)+W<SUB>n+2</SUB>(k-1)X<SUB>n+2</SUB>(k-1)+......,+W<SUB>n+m</SUB>(k-1)X<SUB>n+m</SUB>(k-1)</P>
<P align=right>(2.71) </P>
<P>由于 u(k-1)=Xn+1(k-1)</P>
<P>故而有</P>
<TABLE cellSpacing=0 cellPadding=0 width="80%" align=center border=0>
<TBODY>
<TR>
<TD width="71%"><IMG height=43 src="11.files/5.3.ht22.gif" width=288
border=0></TD>
<TD width="29%">(2.72)</TD></TR>
<TR>
<TD width="71%"><FONT size=2>從式(2.61)和式(2.72)，則有</FONT></TD>
<TD width="29%"></TD></TR>
<TR>
<TD width="71%"><IMG height=111 src="11.files/5.3.ht23.gif"
width=448 border=0></TD>
<TD width="29%">(2.73)</TD></TR>
<TR>
<TD width="71%"><FONT size=2>從引理的性質(zhì)(2)有</FONT></TD>
<TD width="29%"></TD></TR>
<TR>
<TD width="71%"><IMG height=52 src="11.files/5.3.ht25.gif" width=274
border=0></TD>
<TD width="29%">(2.74)</TD></TR></TBODY></TABLE></TD></TR>
<TR>
<TD width="100%" height=2>
<P>只要證明x<SUP>T</SUP>(k-1)x(k-1)是有界的，就可以證明e(<SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">∞</SPAN>)＝0，也就可以證明定理中的性質(zhì)(2)。
</P>
<P>下面證明<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">‖</SPAN>x(k-1)<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">‖</SPAN>有限。</P>
<P>從對象式(2．52)有關(guān)條件，對象的輸入輸出信號滿足</P>
<TABLE cellSpacing=0 cellPadding=0 width="80%" align=center border=0>
<TBODY>
<TR>
<TD width="71%"><IMG height=33 src="11.files/5.3.ht24.gif" width=288
border=0></TD>
<TD width="29%">(2.75)</TD></TR></TBODY></TABLE></TD></TR>
<TR>
<TD width="100%" height=39>
<P>其中：1<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">≤</SPAN>i<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">≤</SPAN>k；m<SUB>1</SUB><<SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">∞</SPAN>；m<SUB>2</SUB><<SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">∞。</SPAN>
</P>
<P>根據(jù)式(2．52)對象的滿足條件，從式(2．53)則有</P>
<TABLE cellSpacing=0 cellPadding=0 width="80%" align=center border=0>
<TBODY>
<TR>
<TD width="71%"><IMG height=36 src="11.files/5.3.ht26.gif" width=512
border=0></TD>
<TD width="29%">(2.76)</TD></TR>
<TR>
<TD width="71%"><FONT size=2>既然，給定信號r是有界的，所以跟蹤誤差有</FONT></TD>
<TD width="29%"></TD></TR>
<TR>
<TD width="71%"><IMG height=31 src="11.files/5.3.ht27.gif" width=364
border=0></TD>
<TD width="29%">(2.77)</TD></TR></TBODY></TABLE></TD></TR>
<TR>
<TD width="100%" height=6>
<P>從而有|e'(k)|+m3<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">≥</SPAN>|y(k)|
</P>
<P>由此，式(2．76)可以寫為：</P>
<TABLE cellSpacing=0 cellPadding=0 width="80%" align=center border=0>
<TBODY>
<TR>
<TD width="86%"><IMG height=42 src="11.files/5.3.ht28.gif" width=701
border=0></TD>
<TD width="14%">(2.78)</TD></TR></TBODY></TABLE></TD></TR>
<TR>
<TD width="100%" height=70>
<P>其中：0<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">≤</SPAN>C<SUB>1</SUB><SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">≤</SPAN><SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">∞</SPAN>；0<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">≤</SPAN>C<SUB>2</SUB><SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">≤</SPAN><SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">∞</SPAN>。
</P>
<P>假設(shè)跟蹤誤差e'(k)有界，則從式(2．78)可知：<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">‖</SPAN>x(k)<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">‖</SPAN>同樣有界；這樣從式(2．74)可知</P>
<TABLE cellSpacing=0 cellPadding=0 width="80%" align=center border=0>
<TBODY>
<TR>
<TD width="71%"><IMG height=35 src="11.files/5.3.ht30.gif" width=128
border=0></TD>
<TD width="29%">(2.79)</TD></TR>
<TR>
<TD width="71%"><FONT size=2>顯然，定理的性質(zhì)(2)成立。
<BR>假設(shè)跟蹤誤差e'(k)無界，則存在時刻序列|kn|，令</FONT></TD>
<TD width="29%"></TD></TR>
<TR>
<TD width="71%"><IMG height=42 src="11.files/5.3.ht29.gif" width=160
border=0></TD>
<TD width="29%">(2.80)</TD></TR></TBODY></TABLE></TD></TR>
<TR>
<TD width="100%">
<P>取m4＝max(1，<SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">ε</SPAN>)
</P>
<P>考慮</P>
<P><IMG height=57 src="11.files/5.3.ht31.gif" width=284 border=0></P>
<TABLE cellSpacing=0 cellPadding=0 width="80%" align=center border=0>
<TBODY>
<TR>
<TD width="70%"><IMG height=247 src="11.files/5.3.ht32.gif"
width=320 border=0></TD>
<TD width="30%">(2.81)</TD></TR>
<TR>
<TD width="70%"><FONT size=2>對式(2.81)取極限有</FONT></TD>
<TD width="30%"></TD></TR>
<TR>
<TD width="70%"><IMG height=159 src="11.files/5.3.ht33.gif"
width=464 border=0></TD>
<TD width="30%">(2.82)</TD></TR></TBODY></TABLE></TD></TR>
<TR>
<TD width="100%" height=1174>
<P>這個極限存在說明e'(K)有界，假設(shè)其無界不成立。 </P>
<P>由于e'(k)有界，故式(2．79)是必定成立的。由于e'(k)＝y(tǒng)(k)-r(k),而r(k)有界,所以，y(k)有界。從式(2．75)可知u(k)也有界。則定理的兩個性質(zhì)成立。</P>
<P>證畢。</P>
<P>四、系統(tǒng)實(shí)際運(yùn)行情況</P>
<P>當(dāng)對象的結(jié)構(gòu)不同時，可以用于檢驗(yàn)圖2-16所示的神經(jīng)適應(yīng)控制系統(tǒng)的運(yùn)行結(jié)果。對象仿真器PE，神經(jīng)控制器NC分別由式(2.52)-(2.55)和式(2.57)-(2.59)所描述；學(xué)習(xí)時采用式(2．56)和式(2．58)。</P>
<P>1．對有噪聲的穩(wěn)定對象的控制</P>
<P>對象由下式表示</P>
<P><IMG height=55 src="11.files/5.3.ht34.gif" width=366 border=0></P>
<P>設(shè)對象仿真器PE和神經(jīng)控制器NC輸入的向量為6個元素，有n＝m＝3。在訓(xùn)練學(xué)習(xí)時PE的權(quán)系數(shù)向量更新取<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">α</SPAN>和<SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">ε</SPAN>的值如下：
<P><IMG height=52 src="11.files/5.3.ht35.gif" width=93 border=0>
<P>權(quán)系數(shù)向量的初始值取
<P>W(0)=[0,0,0,1,0,0]<SUP>T</SUP>
<P align=center><IMG height=342 src="11.files/5.3.ht36.gif" width=467
border=0>
<P align=center>圖2-19 給定值r和對象輸出y
<P align=center><IMG height=253 src="11.files/5.3.ht37.gif" width=461
border=0>
<P align=center>圖2-20 NC產(chǎn)生的控制信號u
<P align=center><IMG height=237 src="11.files/5.3.ht38.gif" width=481
border=0>
<P align=center>圖2-21 PE的學(xué)習(xí)過程W(k)的變化 </P></TD></TR>
<TR>
<TD width="100%" height=1319>
<P>噪聲是平均值為零的高斯白噪聲。 </P>
<P>給定輸入r是幅值為1的方波；每方波周期采樣80次。</P>
<P>控制結(jié)果和情況如圖2—19和圖2—20所示。其中圖2—19是對象輸出和給定值的情況；圖2—20是NC產(chǎn)生的控制信號u(k)。<BR>很明顯，對象仿真器能正確地預(yù)測對象的動態(tài)過程。</P>
<P>圖2—21給出了對象仿真器PE的學(xué)習(xí)過程。</P>
<P>2．對不穩(wěn)定對象的控制</P>
<P>不穩(wěn)定對象由下式表示</P>
<P><IMG height=54 src="11.files/5.3.ht39.gif" width=390 border=0></P>
<P>在系統(tǒng)中，PE和NC的輸入都采用6個元素的向量，故n＝m＝3。在訓(xùn)練學(xué)習(xí)時．PE權(quán)系數(shù)向量更新取<SPAN
style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: 'Times New Roman'; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">α</SPAN>和<SPAN
style="FONT-FAMILY: 宋體; mso-bidi-font-size: 10.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">ε</SPAN>的值為</P>
<P><IMG height=51 src="11.files/5.3.ht40.gif" width=94 border=0></P>
<P>權(quán)系數(shù)向量初始化取值為</P>
<P>W(0)=[0,0,0,1,0,0]<SUP>T</SUP></P>
<P>給定輸入r為幅度為1的方波，方波每周期采樣80次。</P>
<P>控制情況和結(jié)果以及邢學(xué)習(xí)時的w(k)變化情況分別如圖2—22，圖2—23，圖2—24所示。對于不穩(wěn)定對象，顯然在過渡過程中有較大的超調(diào)；但在PE學(xué)習(xí)之后，對象輸出能跟蹤給定r。</P>
<P align=center><IMG height=355 src="11.files/5.3.ht41.gif" width=480
border=0></P>
<P align=center>圖2-22 給定r和對象輸出y的波形</P>
<P align=center><IMG height=238 src="11.files/5.3.ht42.gif" width=457
border=0></P>
<P align=center>圖2-23 NC產(chǎn)生的控制信號U的波形</P>
<P align=center><IMG height=324 src="11.files/5.3.ht43.gif" width=472
border=0></P>
<P align=center>圖2-24 PE學(xué)習(xí)時W(k)的變化情況 </P></TD></TR>
<TR>
<TD width="100%" height=17>
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