采用固相反应法合成了具有硅钛钡石 (Ba2 TiSi2 O8)结构的化合物Sr2 TiSi2 O8,合成温度在12 50℃。对Sr2 TiSi2 O8的XRD衍射图进行指标化表明它属于四方晶系 ,晶胞系数为a =b =0 .83135nm ,c =0 .50 190nm ,α =β =γ =90°。Sr2 TiSi2 O8的粉末倍频信号强度约是KH2 PO4 (KDP)的 8倍。反射光谱显示其紫外吸收边为 2 70nm左右。利用自发成核法进行了Sr2 TiSi2 O8晶体生长研究。 The compound Sr2 TiSi2 O8 with the structure of Ba2 TiSi2 O8 was synthesized by solid state reaction at a temperature of 1250 ℃. The XRD diffraction pattern of Sr2 TiSi2 O8 shows that it belongs to the tetragonal system. The cell coefficients are a = b = 0.83135 nm, c = 0.50 190 nm and α = β = γ = 90 °. The frequency doubling signal strength of Sr2 TiSi2 O8 powder is about 8 times that of KH2 PO4 (KDP). The reflection spectrum shows that the UV absorption edge is about 2 70 nm. The growth of Sr2 TiSi2 O8 crystal was studied by spontaneous nucleation method.